2 Marks Questions - MATHS STD 11 Science Questions - Vidyadip

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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.

A Mathematics book and a Physics book.
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 $= 5 \times 6 = 30$
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,
$3 \times 6 \times 9 \times 12 \times 15 \times 18$
$=3 \times (3 \times 2) \times (3 \times 3) \times (3 \times 4) \times (3 \times 5) \times (3 \times 6)$
$=3^6\times [2 \times 3 \times 4 \times 5 \times 6]$
$=3^6\times (6!)$

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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 $= 4$ Total 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 $= 5 \times 5 \times 5 \times 5$
$= 5^4$
$= 625$
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.

Selecting a boy among $27$ boys.
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 $27 \times 14 = 378$
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$ ^6P_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 $^5P_5 = 5!$ 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 + 10n - 9n - 90 = 0 $
$\Rightarrow n^2 (n + 10) - 9 (n + 10) = 0$
$ \Rightarrow (n^2- 9) (n + 10) = 0$
$\big[\therefore \text{n} +10 \neq 0\big]$
$\Rightarrow n - 9 = 0 $
$\Rightarrow n = 9$
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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2 Marks Questions - MATHS STD 11 Science Questions - Vidyadip