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Permutations · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 68 questions.

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68 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks
Convert the following products into factorials: 1.3.5.7.9....(2n - 1)
Answer
We have, $1 \times 3 \times 5 \times 7 \times 9 \times.........(2\text{n} - 1)$ $=\frac{\big[1.3.5.7.9.....(2\text{n}-1)\big].\big[2.4.6.8.....(2\text{n}-2)(2\text{n})\big]}{2.4.6.8.....(2\text{n}-2)(2\text{n})}$ $=\frac{\big[1.3.5.7.9.....(2\text{n}-1)\big].\big[2.4.6.8.....(2\text{n}-2)(2\text{n})\big]}{2^\text{n}\big[1.2.3.4.......((\text{n-1})(\text{n}))\big]}$ $=\frac{1.2.3.4.5.6.7.8.......(2\text{n-2})(2\text{n-1})(2\text{n})}{2^\text{n}.\text{n}!}$ $=\frac{(2\text{n})!}{2^\text{n}.!}$ $\therefore 1.3.5.7.9......(2\text{n}-1)=\frac{(2\text{n})!}{2^\text{n}.\text{n!}}$
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Question 22 Marks
Given 7 flags of different colours, how many different signals can be generated if a signal requires the use of two flags, one below the other?
Answer
Since there are 7 flags of different colours, therefore, first flag can be selected in 7 ways. Now, the second flag can be selected from any one of the remaining flags in 6 ways. Hence, by the fundamental principle of multiplication, the number of flag is 7 × 6 = 42
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Question 32 Marks
Find the number of different 4-letter words, with or without meanings, that can be formed from the letters of the word 'NUMBER'.
Answer
Total number of letters = 6 $\therefore$ Total number of words = Number of arrangements of6 letters, taken 4 at a time $=\ ^{6}\text{P}_4$$=\frac{6!}{(6-4)!}$
$=\frac{6!}{2!}$
$=\frac{6\times5\times4\times3\times2\times1}{2!}$
$=360$
Hence, the total number of 4 letter words are 360.
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Question 42 Marks
How many words, with or without meaning, can be formed by using the letters of the word 'TRIANGLE'?
Answer
Total number of letters = 8 $\therefore$ Total number of words = Number of arrangement of 8 letters, taken 8 at a time $=\ ^{8}\text{P}_8$$=\frac{8!}{(8-8)!}$
$=\frac{5!}{0!}$
$=8!\ [\because 0!=1]$
Hence, total number of words are 8!
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Question 52 Marks
A number lock on a suitcase has 3 wheels each labelled with ten digits 0 to 9. If opening of the lock is a particular sequence of three digits with no repeats, how many such sequences will be possible? Also, find the number of unsuccessful attempts to open the lock.
Answer
Total number of digits = 10 The digits is not repeats in a sequence of three digits. $\therefore$ Required number of sequences = 10 × 9 × 8 = 720 $\therefore$ Total number of unsuccessful attempts = 720 - 1 = 719
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Question 62 Marks
Find the number of ways in which 8 distinct toys can be distributed among 5 childrens.
Answer
Total numbers of toys $=8$ Total number of children $=5 \therefore$ The total number ways in which 8 distinct toys can be distributed among 5 children. $=5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5=5^8$
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Question 72 Marks
How many words, with or without meaning, can be formed by using all the letters of the word 'DELHI', using each letter exactly once?
Answer
Total number of letters = 5 $\therefore$ Total number of words = Number of arrangement of 5 letters, taken 5 at a time $=\ ^{5}\text{P}_5$$=\frac{5!}{(5-5)!}$
$=\frac{5!}{0!}$
$=5!\ [\because 0!=1]$
$={5\times4\times3\times2\times1}$
$=120$
Hence, the number of words are 120.
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Question 82 Marks
Serial numbers for an item produced in a factory are to be made using two letters followed by four digits (0 to 9). If the letters are to be taken from six letters of English alphabet without repetition and the digits are also not repeated in a serial number, how many serial numbers are possible?
Answer
Each serial number of the product consists of six components. First two are letters and remaining four are numbers. So all the serial numbers will look as shown below.
L
L
N
N
N
N
For the first position of serial number we can have one of the 6 letters. As repetition is not allowed first position of serial number we can have one of the 5 letters. For the third position of serial number we can have one of the 10 numbers. Similarly for the remaining position we can have 9, 8 and 7 possible ways.
L
L
N
N
N
N
$\Uparrow$
$\Uparrow$
$\Uparrow$
$\Uparrow$
$\Uparrow$
$\Uparrow$
6
5
10
9
8
7
So the required number of serial number is 6 × 5 × 10 × 9 × 8 × 7.
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Question 92 Marks
Evaluate the following: $^8\text{P}_3$
Answer
We have, $^8\text{P}_3 = \frac{8!}{(8-3)!}\Big[\because\ ^\text{n}\text{P}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$ $= \frac{8\times7\times6\times5}{5!}$ $= 336$ Hence, $^8\text{P}_3= 336$
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Question 102 Marks
How many three-digit numbers are there with no digit repeated?
Answer
We have to form all possible 3 digit numbers with distinct digits.we cannot have 0 at the hundred's place. so, the hundred's place can be filled with any of the 9 digits 1, 2, 3, 4 .... , 9.
so, there are 9 ways of filling the hundred's place.
Now, 9 digits are left including 0, so, ten's place can be filled with any of the remaining 9 digits in 9 ways. now, the unit's place can be filled which in any of the remaining 8 digits. so, th ere are 8 ways of filling the unit's place.
Hence, the total number of required numbers = 9 × 9 × 8 = 648
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Question 112 Marks
Compute: $\frac{30!}{28!}$
Answer
We have, $\frac{30!}{28!}=\frac{30\times29\times28!}{28!}$$= 30 \times 29$
$= 870$
Hence, $ \frac{30!}{28!}=870$
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Question 122 Marks
From among the 36 teachers in a college, one principal, one vice-principal and the teacher-incharge are to be appointed. In how many ways can this be done?
Answer
Clearly the num bar of ways to appoint one principal, one vice-principal and the teacher- incharge is equal to the number of ways to select the three teachers from the 36 teachers.$\therefore$ Number of ways to appointed 3 teachers = 36 × 35 × 34 = 42840
Hence, the number of ways to appoint one principal, one vice-principal and the teacher-incharge is equal to 42840.
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Question 132 Marks
There are 5 books on Mathematics and 6 books on Physics in a book shop. In how many ways can a students buy.
  1. A Mathematics book and a Physics book.
  2. Either a Mathematics book or a Physics book?
Answer
There are 5 books on mathematics and 6 books on physics in a book shop. The number of ways to select a mathematics book = 5 The number of ways to select a physics book = 6 Now, Number of ways in which a student can buy am a them a tics book and a physics book = 5 × 6 = 30 Number of ways in which a student buy either a mathematics book or a physics book = 5 + 6 = 11
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Question 142 Marks
Convert the following products into factorials: 3.6.9.12.15.18
Answer
We have, $3 \times 6 \times 9 \times 12 \times 15 \times 18=3 \times(3 \times 2) \times(3 \times 3) \times(3 \times 4) \times(3 \times 5) \times(3 \times 6)=3^6 \times[2 \times 3 \times 4 \times 5 \times 6]=3^6$ × (6!)
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Question 152 Marks
In how many ways can five children stand in a queue?
Answer
The total number of ways = Number of arrangements of 5 things, taken all at a time $=^5\text{P}_5$ $=\frac{5!}{(5-5)!}$ $=\frac{5\times4\times3\times2\times1}{0!}\ [\because 0!=1]$ $=120$ Hence, the total numbar of ways in which children stand in a queue is 120.
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Question 162 Marks
How many 9-digit numbers of different digits can be formed?
Answer
In a nine-digit number 0 cannot appear in the first digit. So, the number of ways of filling up the first digit = 9. Now, 9 digits are left including 0. So, second digit can be filled with any of the remaining 9 digits in 9 ways. Similarly, remaining digits can be filled in 8, 7, 6, 5, 4, 3 and 2 ways. Hence, the total number of required numbers = 9 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 = 9 × (9!)
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Question 172 Marks
Compute: $\frac{11!-10!}{9!}$
Answer
We have, $\frac{11!-10!}{9!}=\frac{11\times10\times9!-10\times9!}{9!}$$=\frac{9!\times10[11-1]}{9!}$
$= 10\times10$$=100$
Hence,$ \ \frac{11!-10!}{9!}=100$
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Question 182 Marks
Evaluate the following: $^6\text{P}_6$
Answer
We have, $^6\text{P}_6 = \frac{6!}{(6-6)!}\Big[\because^\text{n}\text{P}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$ $=\frac{6!}{0!}$ $= \frac{6\times5\times4\times3\times2\times1}{1}$ $= 720$ Hence, $^6\text{P}_6= 720$
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Question 192 Marks
If $^\text{n}\text{P}_4= 360,$ find the value of n.
Answer
We have,$^\text{n}\text{P}_4= 360,$
$\Rightarrow \frac{\text{n!}}{(\text{n}-4)!}=360$]
$\Rightarrow \frac{\text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)(\text{n}-4)!=360}{(\text{n}-4)!}$ $\Rightarrow \text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)=6\times 5\times4\times3 $ $\Rightarrow \text{n}=6 \ [13 \text{ by} \text{ comapring}]$ Hence, $\text{n}=6$
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Question 202 Marks
Convert the following products into factorials: 5.6.7.8.9.10
Answer
We have,$5 \times 6 \times 7 \times 8 \times 9 \times 10$
$=\frac{10\times9\times8\times7\times6\times5\times(4\times3\times2\times1)}{4\times3\times2\times1}$ $=\frac{10!}{4!}$ Hence, $5\times6\times7\times8\times9\times10=\frac{10!}{4!}$
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Question 212 Marks
Find the total number of ways in which 20 balls can be put into 5 boxes so that first box contains just one ball.
Answer
Total numbers of balls = 20 Total number of boxes = 5 One ball can be put in first box in 20 ways because we can put any one of the twenty balls in first box. Now, remaining 19 balls are to be but into remaining 4 boxes. This can be done in $4^{19}$ ways; because there are 4 choices for each ball Hence, the required number of ways = $20 \times 4^{19}$.
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Question 222 Marks
How many three-digit odd numbers are there?
Answer
The three digit numbers are 100 to 999 inclusive so there are 999 - 100 + 1 = 999 - 99 = 900 So, 900 three digit numbers If half of all numbers is odd then half of 900 is 450, there are 450 odd positive 3 digit numbers.
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Question 232 Marks
How many three-digit numbers are there?
Answer
We cannot have a Oat the hundred's place. So, the hundred's place can be filled with any of the 9 digits 1, 2, 3 ..... , 9. So, there are 9 ways of filling the hundred's place. Ten's place can be filled with any 10 digits in 10 ways. Now, the unit's place can be filled with any 10 digits in 10 ways. Hence, the total number of required numbers = 9 × 10 × 10 = 900
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Question 242 Marks
There are 6 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next three have 2 each?
Answer
The total number of ways to make attempt to open the lock = 10 × 10 × 10 = 1000 The number of ways to successfuly open the lock = 1 $\therefore $ The number of ways to make an unsuccessful attempt to open the lock = 1000 - 1 = 999. Hence, required number of ways to make an unsuccessfuly attempt to the open the lock is 999.
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Question 252 Marks
How many three-digit numbers are there, with no digit repeated?
Answer
Total number of digits = 10 Total number of 3 digit numbers $= \ ^{10}\text{P}_3 $ But these arrangements also indude those numbers which have O at hundred's place. such numbers are not 3-digit numbers. When 0 is fixed at hundred's place, we have to arrange remaining 9 digits by taking 2 at a time. The number of such arrangements is $= \ ^9\text{P}_3$ So, the total of numbers having O at hundred's place $= \ ^9\text{P}_2$ Hence, total number of 3 digit numbers which distinct $= \ ^{10}\text{P}_3 - ^{9}\text{P}_2 $ $=\frac{10!}{(10-3)!}-\frac{9!}{(9-2)!}$ $=\frac{10!}{7!}-\frac{9!}{7!}$ $=\frac{10\times9\times8\times7!}{7!}-\frac{9\times8\times7!}{7!}$ $= 720-72$ $=648$
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Question 262 Marks
There are 6 items in column A and 6 items in column B. A student is asked to match each item in column A with an item in column B. How many possible, correct or incorrect, answers are there to this question?
Answer
There are 6 items in column A and 6 items in column B Now, Each answer to the given question is an arrangement of the 6 items of column B keeping the order of items in column A fixed. Hence, the total number of answers = Number of arrangements of 6 items in column B $= \ ^6\text{P}_6 $$=\frac{6!}{(6-6)!}$
$=\frac{6!}{0!}$ $={6\times5\times4\times3\times2\times1} \ [\because 0!=1]$ $= 720$
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Question 272 Marks
There are four parcels and five post-offices. In how many different ways can the parcels be sent by registered post?
Answer
Total number of parcels = 4Total number of post-offices = 5
Since a percel can be sent to any one of the five post offices.
So, the requied number of ways = 5 × 5 × 5 × 5
$= 5^4$
$= 625$
Hence, total number of ways is 625.
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Question 282 Marks
In how many ways can six persons be seated in a row?
Answer
First person can be seated in a row in 6 ways. Second person can be seated in a row in 5 ways. Third person can be seated in a row in 4 ways. Fourth person can be seated in a row in 3 ways. Fifth person can be seated in a row in 2 ways. And, sixth person can be seated in a row in 1 ways. Hence, total number of ways in which six persons can be seated in a row = 6 × 5 × 4 × 3 × 2 × 1 = 720.
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Question 292 Marks
In a class there are 27 boys and 14 girls. The teacher wants to select 1 boy and 1 girl to represent the class in a function. In how many ways can the teacher make this selection?
Answer
Here the teacher is to perform two jobs.
  1. Selecting a boy among 27 boys.
  2. Selecting a girl among 14 boys.
The first of these can be perfomed in 27 ways and the second in 14 ways. Therefore by the fumdam pricipal of multiplication, the required number of ways is 27 × 14 = 378 Hence, the teacher can make the selection of a boy a girl in 378 ways.
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Question 302 Marks
In how many ways can an examinee answer a set of ten true/ false type questions?
Answer
The number of ways to ex am i nee answer a true/ false type question is 2. $\therefore$ The number of ways for an ex am i nee to answer a set of ten true/ false type questions = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 1024 Hence, the required number of ways is 1024.
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Question 312 Marks
A coin is tossed five times and outcomes are recorded. How many possible outcomes are there?
Answer
Since toss of each coin can result in 2 ways. When coin is tossed five times, the to outcomes = 2 × 2 × 2 × 2 × 2 = 32 Hence, required number of ways is 32.
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Question 322 Marks
Which of the following are ture. (2 + 3)! = 2! + 3!
Answer
We have, L.H.S. = (2 + 3)! = 5! = 5 × 4 × 3 × 2 × 1 = 120 and, R.H.S. = 2! + 3! = 2 × 1 + 3 × 2 = 2 × 1 + 3 × 2 × 1 = 2 + 6 = 8 $\therefore 120 \neq 8$ $\therefore (2 +3)\neq 2! + 3!$ so it is false
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Question 332 Marks
How many 3-digit numbers are there, with distinct digits, with each digit odd?
Answer
The odd digits are 1, 3, 5, 7, 9 $\therefore$ Total number of odd diqits = 5 Clearly, the hundred's place can be filled with any of the 5 digits 1, 3, 5, 7 or 9 So, there are 5 ways of filling the hundred's place. Now, 4 digits are left. So, ten's place can be filled with any of the remaining 4 digits in 4 ways. Now, the unit's place can be filled with in any of the remaining 3 digits. So, there are 3 ways of filling the unit's place. Hence, the total number of required number = 5 × 4 × 3 = 60
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Question 342 Marks
How many words can be formed from the letters of the word 'SUNDAY'? How many of these begin with D?
Answer
There are 6 letters in the word 'SUNDAY'. The total number of words formed with these 6 letters is the number of arrangements of 6 items, taken all at a time, which is equal to ${ }^6 P _6=6!=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$. If we fix up Din the beginning, then the remaining 5 letters can be arranged in ${ }^5 P_5=5!$ ways. so, the total number of words which begin with $D=5!=5 \times 4 \times 3 \times 2 \times 1=120$.
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Question 352 Marks
If three six faced die each marked with numbers 1 to 6 on six faces, are thrown find the total number of possible outcomes.
Answer
Total numbers of faces in each die = 6 The total number of possible outcomes of three six faced die =6 × 6 × 6 = 216
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Question 362 Marks
There are two works each of 3 volumes and two works each of 2 volumes; In how many ways can the 10 books be placed on a shelf so that the volumes of the same work are not separated?
Answer
Let, $w_1, w_2, w_3$ and $w_4$ be 4 words, where $w_1, w_2$ have 3 volumes each and $w_3, w_4$ have 2 volume each. These 4 works can be arranged in 4 ! ways. Now, volumes of $w_1$ can be arranged in 3 ! ways. volumes of $w_2$ can be arranged in 3 ! ways. volumes of $w_3$ can be arranged in 2 ! ways. And volumes of $w_4$ can be arranged in $2!$ ways. $\therefore$ Total number of ways to arrange all books $=4!(3!\times 3!\times 2!\times 2!)$ $=21 \times 6 \times 6 \times 2 \times 2=3456$
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Question 372 Marks
From among the 36 teachers in a school, one principal and one vice-principal are to be appointed. In how many ways can this be done?
Answer
The total number of teachers in a school = 36 One principal and one uice-principal are to be appointed. $\therefore$ Total of ways = Number of arrangement of 36 things taken two at a time $=\ ^{36}\text{P}_2$$=\frac{36!}{(36-2)!}$
$=\frac{36!}{34!}$
$=\frac{36\times35\times34!}{34!}$
$=36\times35$
$=1260$
Hence, Total number of ways to appoint one principal and one vice-principal are 1260.
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Question 382 Marks
Twelve students complete in a race. In how many ways first three prizes be given?
Answer
Clearly, the total number of ways to select first three prizes is equal to the 3 students from 12 students. $\therefore$ Number of ways to select the three prizes = 12 × 11 × 10 = 1320
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Question 392 Marks
Prove thart. $\frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!}=\frac{122}{11!}$
Answer
$\text{L.H.S.}= \frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!}$$=\frac{1}{9!}+\frac{1}{10\times9!}+\frac{1}{11\times10\times9!}$
$=\frac{11\times10+11+1}{11\times10\times9!}$
$=\frac{110+11+1}{11!}$
$=\frac{122}{11!}$
$\text{R.H.S.}$
Hence, $\frac{1}{9!}+\frac{1}{10!}+\frac{1}{11!}= \frac{122}{11!}$
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Question 402 Marks
How many words can be formed out of the letters of the word, 'ORIENTAL', so that the vowels always occupy the odd places?
Answer
There are 4 vowels and 4 consonants in the word 'ORIENTAL'. We have to arrange 8 leeters in a row such that vowels occupy odd places. There are 4 odd places (1, 3, 5, 7). Four vowels can be arranged in these 4 odd places in 4! ways. Remaining 4 even places (2, 4, 6, 8) are to be occupied by the 4 consonants. This can be done in 4! ways. Hence, the total number of words in which vowels occupy odd places = 4! × 4! = 4 × 3 × 2 × 1 × 4 × 3 × 2 × 1 = 576.
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Question 412 Marks
Convert the following products into factorials: (n + 1)(n + 1)(n + 1) ... (2n)
Answer
We have, (n + 1)(n + 1)(n + 1)...(2n) $=\big[1\times2\times3\times4\times....(\text{n}-1)\text{n}\big]\$\text{n}+2)..(2\text{n}-1)\times2\text{n}\big[1\times2\times3\times4..........(\text{n}-1)\text{n}\big]$ $=\frac{(2\text{n!})!}{\text{n!}}$
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Question 422 Marks
Evaluate the following: $^{10}\text{P}_3$
Answer
We have, $^{10}\text{P}_4 = \frac{10!}{(10-4)!}\Big[\because\ ^\text{n}\text{P}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$$=\frac{10!}{6!}$
$= \frac{10\times9\times8\times7 \times6!}{6!}$ $= 5040$ Hence, $^{10}\text{P}_4= 5040$
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Question 432 Marks
How many five digit telephone numbers can be constructed using the digits 0 to 9. If each number starts with 67 and no digit appears more than once?
Answer
Total number of digits = 10 each number starts with 67 and no digit appears more than once. $\therefore$ Total number of five digit telephone numbers. = 1 × 1 × 8 × 7 × 6 = 336
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Question 442 Marks
Find x in the following. $\frac{\text{x}}{10!}=\frac{1}{8!}+\frac{1}{9!}$
Answer
$\frac{\text{x}}{10!}=\frac{1}{8!}+\frac{1}{9!}$ $\Rightarrow\text{x}=\frac{10!}{8!}+\frac{10!}{9!}$ $\Rightarrow=\frac{10\times9\times8!}{8!}+\frac{10\times9!}{9!}$ $\Rightarrow\text{x}=10\times9+10$ $\Rightarrow\text{x}=100$
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Question 452 Marks
Find x in the following. $\frac{1}{6!}+\frac{1}{7!}=\frac{\text{x}}{8!}$
Answer
$\frac{1}{6!}+\frac{1}{7!}=\frac{\text{x}}{8!}$ $\Rightarrow\text{x}=\frac{8!}{6!}+\frac{8!}{7!}$ $\Rightarrow=\frac{8\times7\times6!}{6!}+\frac{8\times7!}{7!}$ $\Rightarrow\text{x}=8\times7+8$ $\Rightarrow\text{x}=64$
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Question 462 Marks
How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?
Answer
There are 9 ways to pick the 1st digit. For each of those 9 ways there are 8 ways to choose the second digit. That's 9 × 8 or 72 ways to pick the first two digits. For each of those 72 ways there are 7 ways to choose the third digit. That's 72 × 7 ways or 504 ways to pick all three digits.
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Question 472 Marks
How many numbers of four digits can be formed with the digits 1, 2, 3, 4, 5 if the digits can be repeated in the same number?
Answer
Total number of digits = 5 Since, the digits can be repeated in the same number. Total numbers of four digits numbers = 5 × 5 × 5 × 5 = 625
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Question 482 Marks
From Goa to Bombay there are two roots; air, and sea. From Bombay to Delhi there are three routes; air, rail and road. From Goa to Delhi via Bombay, how many kinds of routes are there?
Answer
From Goa to Bombay there are two roots; air and sea. From Bombay to Delhi there are three routs; air rail and raod. Therefore by the fundamental pricipal of multiplication, the requied number fo ways are 2 × 3 = 6 Hence, total number of different kinds routes are 6.
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Question 492 Marks
If $(n + 1)! = 90 [(n - 1)!]$, find $n$.
Answer
We have, $(n+1)!=90[(n-1)!]$
$\Rightarrow(n+1)!\times n \times(n-1)!=90[(n-1)!]$
$\Rightarrow n(n+1)=90$
$\Rightarrow n^2+10 n-9 n-90=0$
$\Rightarrow n^2(n+10)-9(n+10)=0$
$\Rightarrow\left(n^2\right.$
$-9)(n+10)=0[\therefore n+10 \neq 0]$
$\Rightarrow n-9=0$
$\Rightarrow n=9 \text { Hence, } n=9$
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Question 502 Marks
In how many ways can 5 different balls be distributed among three boxes?
Answer
Total number of balls = 5 Total number of boxes = 3 $\therefore$ Total number of ways to distributed 5 different balls in three boxes = 3 × 3 × 3 × 3 × 3 = 243
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Question 512 Marks
How many words can be formed from the letters of the word 'SUNDAY'? How many of these begin with D?
Answer
There are 6 letters in the word 'SUNDAY'. The total number of words formed with these 6 letters is the number of arrangements of 6 items, taken all at a time, which is equal to ${ }^6 P _6=6!=6 \times 5 \times 4 \times 3 \times 2 \times 1=720$. If we fix up Din the beginning, then the remaining 5 letters can be arranged in ${ }^5 P_5=5$ ! ways. so, the total number of words which begin with $D=5!$
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Question 522 Marks
How many 6-digit telephone numbers can be constructed with digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 if each number starts with 35 and no digit appears more than once?
Answer
Total number of digits = 10 The first two digits of telephone is 35 and no digit appears more than once.$\therefore$ Total number of remaining digits = 10 - 2 = 8
And, Total number of remaining digits of telephone number = 6 - 2 = 4 $\therefore$ Required number of telephone numbers $=\ ^8\text{P}_4$ $=\frac{8!}{(8-4)!}$ $=\frac{8!}{4!}$ $=\frac{8\times7\times6\times5\times4!}{4!}$ $= 1680$
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Question 532 Marks
How many words can be formed out of the letters of the word 'ARTICLE', so that vowels occupy even places?
Answer
We have to arrange 7 letters in a row such that vowels occupy even places. There are 3 even places (2, 4, 6). Three vowels can be arranged in these 3 even places in 3! ways. Remaining 4 odd places (1, 3, 5, 7) are to be occupied by the 4 consonants. This can be done in 4! ways. Hence, the total number of words in which vowels occupy even places = 3! × 4! = 3 × 2 × 4 × 3 × 2 = 144
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Question 542 Marks
There are 6 multiple choice questions in an examination. How many sequences of answers are possible, if the first three questions have 4 choices each and the next three have 2 each?
Answer
Each one of the first three questions can be answered in 4 ways. $\therefore$ The total number of ways to answered the first three question = 4 × 4 × 4 = 64 Each of the next three question can be answered in 2 ways. $\therefore$ The total number of ways the answered the next three qusstions = 2 × 2 × 2 = 8so, total number of sequences at answers = 64 × 8 = 512
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Question 552 Marks
How many numbers of six digits can be formed from the digits 0, 1, 3, 5, 7 and 9 when no digit is repeated? How many of them are divisible by 10?
Answer
Total number of digits = 6 we cannot have Oat the first digit of the required six-digit numbers The digits cannot repeat in the six digits number. $\therefore$ Total number of six digit number are = 5 × 5 × 4 × 3 × 2 × 1 = 600 Now, the six digit number can be divided by 10, if its last digit is 0. $\therefore$ Total numbers which are divisible by 10 = 5 × 4 × 3 × 2 × 1 × 1 = 120
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Question 562 Marks
A team consists of 6 boys and 4 girls and other has 5 boys and 3 girls. How many single matches can be arranged between the two teams when a boy plays against a boy and a girl plays against a girl?
Answer
A boy can be selected from the first team in 6 ways, and from the second in 5 ways. so, number of single matches between the boys of two teams = 6 × 5 = 30. similarly, the number of single matches between the girls of two teams = 4 × 3 = 12. so, total number of matches = 30 + 12 = 42.
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Question 572 Marks
How many A.P.'s with 10 terms are there whose first term is in the set {1, 2, 3} and whose common difference is in the set {1, 2, 3, 4, 5}?
Answer
There are 3 ways to choose the first form and corresponding to each such way there are 5 ways of selecting the comm on difference. So, required number of A.P.'s = 3 × 5 = 15
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Question 582 Marks
In how many ways can 6 boys and 5 girls be arranged for a group photograph if the girls are to sit on chairs in a row and the boys are to stand in a row behind them?
Answer
Total number of boys = 6 Total number of girls = 5 Now, Five girls can sit on chairs in a row in $^5P_5 = 5!$ ways. and 6 boys can stand behind them in a row in $^6P_6 = 6!$ ways. Hence, the total number of ways = 5! × 6! = 5 × 4 × 3 × 2 × 1 × 6 × 5 × 4 × 3 × 2 × 1 = 120 × 720 = 86400
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Question 592 Marks
A person wants to buy one fountain pen, one ball pen and one pencil from a stationery shop. If there are 10 fountain pen varieties, 12 ball pen varieties and 5 pencil varieties, in how many ways can he select these articles?
Answer
Here the person is to perform three jobs.
  1. Selection a ball pen from 12 ball pens.
  2. Selection a fountain pen form 10 fountain pens.
  3. Selection a pencil form 5 pencils.
The first of these can be perfomed in 12 ways, the second in 10 ways and thrid in 5 ways. Therefore by the fundamental pricipal of multipilcation, the requied number of ways is 12 × 10 × 5 = 600 Hence, the person can make the selection of a fountain pen, ball pen and pencil in 600 ways.
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Question 602 Marks
Four books, one each in Chemistry, Physics, Biology and Mathematics, are to be arranged in a shelf. In how many ways can this be done?
Answer
Total number of books = 4 $\therefore$ Total number of ways = Number of arrangments of 4 books, taken all at a time $= \ ^{4}\text{P}_4$$=\frac{4!}{(4-4)!} \ \bigg[\because ^\text{n}\text{p}_\text{r}=\frac{\text{n!}}{(\text{n-r})!}\bigg]$
$=\frac{4!}{0!}$
$=4! \ [\because 0!=1]$
$=4\times3\times2\times1$
$=24$
Hence, the total number of ways to arrange the books in a shelf = 24
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Question 612 Marks
Find x in the following. $\frac{1}{4!}+\frac{1}{5!}=\frac{\text{x}}{6!}$
Answer
We have, $\frac{1}{4!}+\frac{1}{5!}=\frac{\text{x}}{6!}$ $\Rightarrow​​​​​​​​\frac{1}{4!}+\frac{1}{4\times5!}=\frac{\text{x}}{6\times5\times4!}$ $\Rightarrow4!\times\bigg[\frac{1}{4!}+\frac{1}{5\times4!}\bigg]=\frac{\text{x}}{30}$ $\Rightarrow1+\frac{1}{58}=\frac{\text{x}}{30}$ $\Rightarrow\frac{6}{5}=\frac{\text{x}}{30}$ $\Rightarrow\frac{\text{x}}{30}=\frac{6}{5}$ $\Rightarrow\text{x}=\frac{6\times30}{5}$ $\Rightarrow6\times6$ $\Rightarrow\text{x}= 36$ Hence, $\text{x}= 36$
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Question 622 Marks
How many different numbers of six digits can be formed from the digits 3, 1, 7, 0, 9, 5 when repetition of digits is not allowed?
Answer
We cannot have 0 at the first dig it of six-digit numbers. So, the first digit of six-digit numbers can be selected in 5 ways. Now, 5 digits are left including 0. So, second digit of six-digit numbers can be selected in 5 ways. Third digit of six-diqit numbers can be selected in 4 ways. Fourth digit of six-digit numbers can be selected in 3 ways Fifth digit of six-digit numbers can be selected in 2 ways. Last digit of six-digit numbers can be selected in 1 ways. Hence, total number of num bers = 5 × 5 × 4 × 3 × 2 × 1 = 600
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Question 632 Marks
How many different five-digit number licence plates can be made if.
  1. First digit cannot be zero and the repetition of digits is not allowed.
  2. The first-digit cannot be zero, but the repetition of digits is allowed?
Answer
  1. Zero cannot be first digit of the license plates.
This means the first digit can be selected from the 9 digits 1, 2, 3, 4 .... , 9 So, there are 9 ways of filling the first digit of the license plates.
Now, 9 digits are Ieft including 0. So, second place can be filled with any of the remaining 9 digits in 9 ways.
The third place of the license plates can be filled with in any of the remaining 8 digits. So, there are 8 ways of filling the third place.
The fourth place of the license plates can be filled with in any of the remaining 7 digits. So, there are 7 ways at filling the fourth place.
The last place of the license plates can be filled with in any of the remaining 6 digits. So, there are 6 ways of filling the fourth place.
Hence, the total number of ways = 9 × 9 × 8 × 7 × 6 = 27216
  1. Zero cannot be first digit of the license plates.
$\therefore$ First digit can be selected from the 9 digits 1, 2, 3 ..... , 9 So, there are 9 ways at filling the first digit of the licence plates.
The repetition of digits is allowed to made a license plates number.
$\therefore$ The number of ways to fill the remaining places of the number platas = 10 × 10 × 10 × 10.
Hence, the total number of ways = 9 × 10 × 10 × 10 × 10 = 90,000.
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Question 642 Marks
A mint prepares metallic calendars specifying months, dates and days in the form of monthly sheets (one plate for each month). How many types of calendars should it prepare to serve for all the possibilities in future years?
Answer
The mint has to perform two job,
  1. Selecting the number of days in the february month (there can be 28 days or 29 days).
  2. Selecting the first day of february.
The first job can be compeleted in 2 ways the second can be performed in 7 ways by selecting are on of the seven days of a week. Thus, the required number of plates = 2 × 7 = 14 Hence, total number of calendars = 7 × 2 = 14
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Question 652 Marks
Evaluate the following:P(6,4)
Answer
We have, ${P}(6,4) = \frac{6!}{(6-4)!}\Big[\because^\text{n}\text{P}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$ $=\frac{6!}{2!}$ $= \frac{6\times5\times4\times3\times2\times1}{2!}$ $= 360$ Hence, $\text{P}(6,4)= 720$
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Question 662 Marks
How many four digit different numbers, greater than 5000 can be formed with the digits 1, 2, 5, 9, 0 when repetition of digits is not allowed?
Answer
Since the required num bars are greater than 5000. $\therefore$ the thousand's place can be filled with any of two digits 5 or 9. So, there are 2 ways of filling the thousand's place. Since repetition of digits is not allowed, so the hundred's ten's and one's places can be filled in 4, 3 and 2 ways respectively. Hence, the required number of num bers = 2 × 4 × 3 × 2 = 49
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Question 672 Marks
In how many ways can the letters of the word 'FAILURE' be arranged so that the consonants may occupy only odd positions?
Answer
There are 4 vowels and 3 consonants in the word 'FAILURE'. We have to arrange 7 letters in a row such that consonants occupy odd places. There are 4 odd places (1, 3, 5, 7). There consonants can be arranged in these 4 odd places in $^4P_3$ ways. Remaining 3 even places (2, 4, 6) are to be occupied by the 4 vowels. This can be done in $^4P_3$ ways. Hence, the total number of words in which consonants occupy odd places $= \ ^4\text{P}_3 \times\ ^4\text{P}_3$ $= \frac{4!}{(4-3)!} \times \frac{4!}{(4-3)!}$ $= 4\times3\times 2\times 1\times 4\times3\times 2\times 1$ $=24\times24$ $=576$
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Question 682 Marks
Compute: L.C.M. (6!,7!,8!)
Answer
We have, 8! = 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 7! = 7 × 6 × 5 × 4 × 3 × 2 × 1 6! = 6 × 5 × 4 × 3 × 2 × 1 and $\therefore$ L.C.M. (6!,7!,8!) = 8!
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