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MATHS 3 Marks Question

Permutations · UP Board (UPMSP) STD 11 Science (UPMSP - English Medium). 49 questions.

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Question 13 Marks
If P(5, r) = P(6, r - 1), find r
Answer
We have,
P(5, r) = P(6, r - 1)
$\Rightarrow \frac{5!}{(5-\text{r})!}= \frac{6!}{\big[6-(\text{r}-1)\big]!}\ \Big[\because^\text{n}\text{p}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$
$\Rightarrow \frac{1}{(5-\text{r})!}=\frac{6}{[7-\text{r}]}!$
$\Rightarrow \frac{1}{(5-\text{r})!}= \frac{6}{(7-\text{r})\times(7-\text{r}-1)(7-\text{r}-2)!}$
$\Rightarrow \frac{1}{(5-\text{r})!}=\frac{6}{(7-\text{r})\times(6-\text{r})(5-\text{r}!)}$
$\Rightarrow 1 = \frac{6}{(7-\text{r})\times(6-\text{r})}$
$\Rightarrow(6-\text{r})\times(7-\text{r})= 6$
$\Rightarrow 42-6\text{r}-7\text{r}+\text{r}^2= 6$
$\Rightarrow \text{r}^2-12\text{r}+42-6=0$
$\Rightarrow \text{r}^2-9\text{r}-4\text{r}+36 = 0$
$\Rightarrow \text{r}(\text{r}-9)-4(\text{r}-9)= 0$
$\Rightarrow (\text{r}-9)(\text{r}-4)= 0$
$\Rightarrow \text{r}= 4 \begin{bmatrix}\because\text{r}\ \leq\ \text{n} \\ \therefore \ \text{r}-9\neq0 \end{bmatrix}$
Hence, $\text{r}= 4$
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Question 23 Marks
How many three digit numbers can be formed by using the digits 0, 1, 3, 5, 7 while each digit may be repeated any number of times?
Answer
Total number of digits = 5
We cannot have 0 at the hundred's place so, the hundred's place can be digits with any of the 4 digits 1, 3, 5 or 7, So, there are 4 ways of filling the hundred's place.
Since, the digit may be repeated in three digit numbers.
$\therefore$ Ten's place can be filled with any of the 5 digits in 5 ways.
And unit's place can be filled with any of the 5 digits in 5 ways.
Hence, the total number of required numbers = 4 × 5 × 5 = 100
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Question 33 Marks
Find the number of 4-digit numbers that can be formed using the digits 1, 2, 3, 4, 5, if no digit is repeated? How many of these will be even?
Answer
We can take the digits one at a time, starting at either end.
Let's start from the right.
d c b a = the digits to be chosen.
For a we have 5 choices (1, 2, 3, 4, 5)
For b we only have 4 (having used one for a, and repeats not allowed)
For c we have 3
Ford we have 2.
5 × 4 × 3 × 2 = 120 choices overall.
If we want the number to be even, we don't have 5 choices for a, we are limited to the set {2, 4} there are only two digits available.
But for the remaining digits the calculation is the same.
$\frac{2}{5}$ of the numbers are even $=\frac{2}{5}\times120=48=2\times4\times3\times2$
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Question 43 Marks
In how many ways can the letters of the word 'ALGEBRA' be arranged without changing the relative order of the vowels and consonants?
Answer
There are 4 consonants in the word 'ALGEBRA'.
The number of ways to arrange these consonants = 4!
There are 3 vowels in the given word of which 2 are A's
The vowels can be arranged among them selves in $\frac{3!}{2!}$ ways.
Hence, the required number of arrangements $=4!\times\frac{3!}{2}$
$=4\times 3\times 2\times \frac{3\times 2}{2}$
$=72$
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Question 53 Marks
In how many ways can $7$ letters be posted in $4$ letter boxes$?$
Answer
Total number of ball $= n = 5$
Number of boxes $= r = 3$
$5$ different balls can be distributed among three boxes in $^5P_3$ ways.
$^5\text{p}_3= \frac{5!}{(5-3)!}= \frac{5!}{2!}=\frac{5\times 4\times 3\times 2\times 1}{2 \times 1}=60.$
In $60$ ways $5$ different balls can be distributed among three boxes.
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Question 63 Marks
If $\frac{\text{(2n!)}}{3!(2\text{n}-3)!}$ and $\frac{\text{n!}}{2!(2\text{n}-2)!}$ are in the ratio 44 : 3 find n.
Answer
We have,
$\frac{\frac{(2\text{n!})}{3!(2\text{n}-3)!}}{\frac{\text{n!}}{2!(\text{n}-2)!}} = \frac{44}{3}$
$\Rightarrow \frac{(2\text{n})! \times2! (\text{n-2})!}{3! (2\text{n}-3)! \times n!} = \frac{44}{3}$
$\Rightarrow \frac{(2\text{n})(2\text{n}-1)(2\text{n-2})(2\text{n-3})!\times2! (\text{n}-2)!}{3\times2(2\text{n-3})\times\text{n}(\text{n}-1)(\text{n-2})}= \frac{44}{3}$
$\Rightarrow \frac{2\text{n}(2\text{n-1})\times2(\text{n-1})}{3(\text{n-1})}=\frac{44}{3}$
$\Rightarrow 4(2\text{n}-1)=44$
$\Rightarrow 2\text{n}-1=11$
$\Rightarrow 2\text{n}=12$
$\Rightarrow \text{n}= 6$
$\therefore \text{n}=6$
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Question 73 Marks
In how many ways can the letters of the word 'ARRANGE' be arranged so that the two R's are never together?
Answer
There are 7 letters in the word 'ARRANGE' out of which 2 are A's 2 are R's and the rest are all distinct. So, total number of words $=\frac{7!}{2!\ 2!}$ $=\frac{7\times6\times5\times4\times3\times2!}{2\times2!}$ $=7\times6\times5\times2\times3$ $=1260$Considering all R's together and treating them as one letter we have 6 letters out of which A repeats 6! 2 times and other are distinct. These 6 letters can be arranged in $\frac{6!}{2!}$ ways.
So, the number of words in which all R's come together $=\frac{6!}{2!}$
$=\frac{6\times5\times4\times3\times2!}{2!} = 360$Hence, the number of words in which all R's do not came together,
= Total number of words - Number of words in which all R's come together
= 1260 - 360
= 900.
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Question 83 Marks
How many numbers can be formed with the digits 1, 2, 3, 4, 3, 2, 1 so that the odd digits always occupy the odd places?
Answer
Total number of digits are = 7
There are 4 odd digits 1, 1, 3, 3 and 4 odd places (1, 3, 5, 7)
So, odd digits can be arranged in odd places in $=\frac{4!}{2!\ 2!}$
The 3! remaining 3 even digits 2, 2, 4 can be arranged in 3 even places in $\frac{3!}{2!}$ ways.
Hence, the total number of Numbers $=\frac{4!}{2!\ 2!}\times\frac{3!}{2!}=\frac{4\times3\times2!}{2!\ 2!}\times\frac{3\times2!}{2!}=18$
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Question 93 Marks
A coin is tossed three times and the outcomes are recorded. How many possible outcomes are there? How many possible outcomes if the coin is tossed four times? Five times$?\ n$ times$?$
Answer
Since a toss of a coin can result in a head or a tail.
$\therefore$ Total number of possible outcomes in each tossed $= 2$
$\therefore$ Total number of possible outcomes in four tossed $= 2 \times 2 \times 2 = 2^3 = 8$
$\therefore$ Total number of possible outcomes in four tossed $= 2 \times 2 \times 2 \times 2 = 2^4 = 16$
$\therefore$ total number of possible outcomes in five tossed $= 2 \times 2 \times 2 \times 2 \times 2 = 2^5 = 32$
$\therefore$ total number of possible outcomes inn tossed $= 2 \times 2 \times 2 ..... n$ times $= 2^n$
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Question 103 Marks
If $(n + 3)! = 56 [(n + 1)!],$ find $n.$
Answer
We have,$(n + 3)! = 56 [(n + 1)!]$
$\Rightarrow (n + 3)! \times (n + 2)! \times (n + 1)! = 56 [(n - 1)!] $
$\Rightarrow (n + 2)(n + 3) = 56 $
$\Rightarrow n^2 + 3n + 2n + 6 = 56$
$\Rightarrow n^2 + 5n + 6 - 56 = 0 $
$\Rightarrow n^2 + 5n + 50 = 0 $
$\Rightarrow n^2 + 10n - 5n - 50 = 0 $
$\Rightarrow n (n + 10) - 5 (n + 10) = 0 $
$\Rightarrow (n - 5)(n + 10) = 0$
$\big[\therefore \text{n} +10 \neq 0\big]$
$\Rightarrow n - 5 = 0$
$​​​​​​​ \Rightarrow n - 5 = 0 $
$\Rightarrow n = 5$
Hence, $n = 5$
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Question 113 Marks
How many numbers greater than 1000000 can be formed by using the digits 1, 2, 0, 2, 4, 2, 4?
Answer
Total number of digits = 7
Now,
number of 7-digit numbers
$=\frac{7!}{3!\ 2!}$
$=\frac{7\times6\times5\times4\times3!}{3!\times2}$
$=7\times6\times5\times2$
$=420$
And, 0 cannot be first digit of the 7-digit numbers
$\therefore$ Number of 6-digit numbers
$=\frac{6!}{3!\ 2!}$
$=\frac{6\times5\times4\times3!}{3!\times2}$
$=6\times5\times2$
$=60$
Hence, total number of 7-digit number which are greater than 1000000 = 420- 60 = 360.
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Question 123 Marks
Three dice are rolled. Find the number of possible outcomes in which at least one die shows 5.
Answer
Total num bars of dice = 3
$\therefore$ The number of possible outcomes
= 6 × 6 × 6 = 216
$\therefore$ Total number of possible outcomes in which S dose not appear on any dice
= 5 × 5 × 5 = 125
$\therefore$ Required number of possible outcomes
= 216 - 125 = 91
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Question 133 Marks
How many words can be formed from the letters of the word 'SERIES' which start with S and end with S?
Answer
In the word 'SERIES' there are 6 letters of which 2 are S and 2 are E's. Let us fix 5 at the extreme left and at the extreme right end. Now, we are left Let us fix 5 at the extreme left and at the extreme right end. Now, we are left $\frac{4!}{2!}$ ways. Hence, required number of arrangements $=\frac{4!}{2!}$ $=\frac{4\times3\times2!}{2!}$$=12$
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Question 143 Marks
How many different words can be formed with the letters of word 'SUNDAY'? How many of the words begin with N? How many begin with $N$ and end in Y?
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 Nin the begining, then the remaining 5 letters can be arranged in $^5P_5 = 5!$ ways so, the total num bar of words which begin which $N = 5!$
$= 5 \times 4 \times 3 \times 2 \times 1$
$= 120$
if we fix up Nin the begining and $Y$ at the end, then the remaining $4$ letters can be arranged in $^4P_4 = 4!$ ways.
So, the total number of words which begin with $N$ and end with $Y = 4! = 4 \times 3 \times 2 \times 1 = 24.$
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Question 153 Marks
A customer forgets a four-digits code for an Automatic Teller Machine (ATM) in a bank. However, he remembers that this code consists of digits 3, 5, 6 and 9. Find the largest possible number of trials necessary to obtain the correct code.
Answer
Total number of digits = 4.
$\therefore$ The largest possible number of trials to obtain the correct code = 4 × 3 × 2 × 1
= 24 [$\because$ digits are not repeated]
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Question 163 Marks
How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 7000, if repetition of digits is not allowed?
Answer
The required numbers are greater than 7000.
$\therefore$ the thousand's place can be filled with any of the 3 digits 7,8,9.
so, there are 3 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 numbers = 3 × 4 × 3 × 2 = 72
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Question 173 Marks
How many four-digit numbers can be formed with the digits 3, 5, 7, 8, 9 which are greater than 8000, if repetition of digits is not allowed?
Answer
Since the required numbers are greater than 8000.
$\therefore$ The thousand's place can be two digits 8 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 numbers = 2 × 4 × 3 × 2 = 48
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Question 183 Marks
How many natural numbers less than 1000 can be formed from the digits 0, 1, 2, 3, 4, 5 when a digit may be repeated any number of times?
Answer
Total number of digits = 6
Clearly, the natural numbers ten's than 1000 can be 3 digits, 2 digits and 1 digit numbers.
Now, 0 cannot be a first dig it of the three digit numbers.
So, the hundred's place can be filled with any of the 5 digits 1, 2, 3 .... 5. So, there are 5 ways of filling the hundred place.
The ten's place can be filled with in any of the 6 digits 0, 1, 2 ..... 5. So, there are 6 ways of filling the ten's place.
The unit's place can be filled with in any of the 6 digits 0, 1, 2 ..... 5. So, there are 6 ways of filling the ten's place.
$\therefore$ The total number of 3 digit numbers = 5 × 6 × 6 = 180
Similarly, the total number of 2 digit numbers = 5 × 6 = 30
Now, 0 is not a natural number
$\therefore$ The total number of ldigit numbers = 5
$\therefore$ Total number of natural numbers tens than 1000
= 180 + 30 + 5 = 215.
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Question 193 Marks
If P(n − 1, 3) : P(n, 4) = 1 : 9, find n.
Answer
We have,
P(n − 1, 3) : P(n, 4) = 1 : 9
$\Rightarrow \frac{\text{p}(\text{n-1,3})}{\text{P} (\text{n}, 4)}=\frac{1}{9}$
$\Rightarrow\frac{\frac{\text{(n-1)!}}{(\text{(n-1-3)!})}}{\frac{\text{n}!}{\text{(n-4)!}}}=\frac{1}{9}$
$\Rightarrow \frac{\text{(n-1) }\times\text{(n-4)!}}{\text{(n-4)!}\times\text{n}!}=\frac{1}{9}$
$\Rightarrow \frac{\text{(n-1)}}{\text{n!}}=\frac{1}{9}$
$\Rightarrow \frac{\text{(n-1)!}}{\text{n}\times(\text{n-1})!}=\frac{1}{9}$
$​​\Rightarrow\frac{1}{\text{n}}=\frac{1}{9}$
$\Rightarrow\text{n}=9$
Hence, $\text{n}=9$
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Question 203 Marks
Which of the following are ture.
(2 × 3)! = 2! × 3!
Answer
We have,
L.H.S. = (2 × 3)!
= 6!
= 6 × 5 × 4 × 3 × 2 × 1
= 720
and, R.H.S.
= 2! × 3!
= 12
$\therefore 720 \neq 12$
$\therefore (2 \times3)\neq 2! \times 3!$
so it is false
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Question 213 Marks
$m$ men and $n$ women are to be seated in a row so that no two women sit together, if $m > n$ then show that the number of ways in which they can be seated as $\frac{\text{m}! (\text{m}+1)!(\text{m}−\text{n}+1)!}{\text{m}! (\text{m}+1)!(\text{m}-\text{n}+1) !}$
Answer
$m$ men can be seated in a row in $^mp_m = m!$ ways.
Now, in the $(m + 1)$ gaps $n$ women can be arranged in $^{m+1}p_n$ ways.
Hence, the number of ways in which no two women sit together.
$= \text{ m!}\times \ ^{\text{m}+1}\text{P}_\text{n}$
$=\text{m}!\times \frac{(\text{m}+1)}{(\text{m}+1-\text{n})!}$
$=\text{m}!\times \frac{(\text{m}+1)}{(\text{m}-\text{n}+1)!}$
Hence, proved.
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Question 223 Marks
In how many ways can $4$ prizes be distributed among $5$ students, when
  1. No student gets more than one prize?
  2. A student may get any number of prizes?
  3. No student gets all the prizes?
Answer
  1. $4$ prizes be distributed among $5$ students so that no student gets more than one prize can be done in
$^5\text{p}_4= \frac{5!}{(5-4)!}=\frac{5!}{(1)!}= 5! \ \text{ways}.$
  1. The first prize can be given away in $5$ ways as it may be given to anyone of the $5$ students. The second prize can also be given away in $5$ ways, since if may be obtained by the student who has already received a prize. Similarly, third and fourth prize can be given away in $5$ ways.
Hence, the number of ways in which all the prize can be given away $= 5 \times 5 \times 5 \times 5 = 625$
  1. Since any of the $5$ students may get all the prizes. So, the number of ways in which a student gets all the $4$ prizes is $5.$
So, the number of ways in which a student does not get all the prizes $= 625 - 5 = 620$
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Question 233 Marks
If 5 P(4, n) = 6.P(5, r - 1), find n.
Answer
We have,
5 P(4, n) = 6.P(5, r - 1)
$\Rightarrow 5\times\frac{4!}{(4-\text{n})!}=6\times \frac{5!}{\big[5-(\text{r}-1)\big]!}\ \Big[\because^\text{n}\text{p}_\text{r}=\frac{\text{n!}}{(\text{n}-\text{r})!}\Big]$
$\Rightarrow5\times \frac{4}{(4-\text{n})!}=\frac{6\times5\times4!}{[5-\text{n}+1]}!$
$\Rightarrow \frac{1}{(4-\text{n})!}= \frac{6}{[6-\text{n}]!}$
$\Rightarrow \frac{1}{(4-\text{n})!}=\frac{6}{(6-\text{n})\times(6-\text{n}-1)(6-\text{n}-2)!}$
$\Rightarrow \frac{1}{(4-\text{n})}= \frac{6}{(6-\text{n})(5-\text{n})(4-\text{n})!}$
$\Rightarrow \frac{(6-\text{n})(5-\text{n})(4-\text{n})!}{(4-\text{n})!}=6$
$\Rightarrow(6-\text{n})\times(5-\text{n})= 6$
$\Rightarrow 30-6\text{n}-5\text{n}+\text{n}^2= 6$
$\Rightarrow \text{n}^2-11\text{n}+30=6$
$\Rightarrow \text{r}^2-11\text{r}+24 = 0$
$\Rightarrow \text{n}^2-8\text{n}-3\text{n}+24=0 $
$\Rightarrow \text{n}(\text{n}-8)-3(\text{n}-8)= 0$
$\Rightarrow (\text{n}-8)(\text{n}-3)= 0$
$\Rightarrow \text{n}-3 = 0$
$\Rightarrow \text{r}= 3 \begin{bmatrix}\because\text{n}\ \leq\ 4 \\ \therefore \ \text{n}\neq8 \end{bmatrix}$
Hence, $\text{r}= 3$
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Question 243 Marks
prove that:
$\frac{(2\text{n+1})}{\text{n!}}=2^\text{n}\bigg\{1.3.5....(2\text{n}-1)(2\text{n}+1)\bigg\}$
Answer
We have,
$\text{L.H.S.}= \frac{(2\text{n+1})}{\text{n!}}$
$=\frac{(2\text{n+1)}\big[1.2.3.4.5.6.7.8.9........(2\text{n-1})\text{2n}\big]}{\text{n!}}$
$=\frac{\big[1.3.5.7....(2\text{n}-1)\times(2\text{n}+1)\big]\big[2.4.6.8.....(2\text{n-2})2\text{n}\big]}{\text{n!}}$
$=\frac{\big[1.3.5.7....(2\text{n-1})(2\text{n}+1)\big]\times 2^\text{n}\big[1.2.3.4.....(\text{n}-1) \text{n}\big]}{\text{n!}}$
$=\frac{\big[1.3.5.7.....(2\text{n}-1)(2\text{n}+1)\big]2^\text{n}\times\text{n!}}{\text{n!}}$
$= 2^\text{n} [ 1.3.5.7.......(2\text{n} - 1)(2\text{n} + 1)]$
$=\text{R.H.S.}$
Hence proved.
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Question 253 Marks
There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated.
Answer
Each lamps has two possibilities either it can be switched on or off.
There are 10 lamps in the hall.
So the total numbers of possibilities are 210.
To illuminate the hall we require at least one lamp is to be switched on.
There is one possibility when all the lamps are switched off.
If all the bulbs are switched off then hall will not be illuminated.
So the number of ways in which the hall can be illuminated is 210 - 1.
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Question 263 Marks
How many different numbers, greater than 50000 can be formed with the digits 0, 1, 1, 5, 9.
Answer
Total number of digits = 5
Now, numbers greater then 50000 will have either 5 or 9 in the first place and will consist of 5 digits.
Number of numbers of which digit 5 at first place $=\frac{4!}{2!}$ [$\because$ 1 is repeated ]
$=\frac{4\times3\times2!}{2!}$
$=12$
Number of numbers with digit 9 at first place $=\frac{4!}{2!}​​​​=12$
Hence, the required number of numbers= 12 + 12 = 24.
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Question 273 Marks
How many different words can be formed with the letters of word 'SUNDAY'? How many of the words begin with $N?$ How many begin with $N$ and end in $Y?$
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 Nin the begining, then the remaining $5$ letters can be arranged in $^5P_5 = 5!$ ways
so, the total num bar of words which begin which $N = 5!$
$= 5 \times 4 \times 3 \times 2 \times 1$
$= 120.$
if we fix up Nin the begining and $Y$ at the end, then the remaining $4$ letters can be arranged in $^4P_4 = 4!$ ways.
So, the total number of words which begin with $N$ and end with $Y = 4! = 4 \times 3 \times 2 \times 1 = 24.$
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Question 283 Marks
If (n + 2)! = 60 [(n - 1)!], find n.
Answer
We have,(n + 2)! = 60 [(n - 1)!]
(n + 2) (n + 1) (n) (n - 1)! = 60 [(n - 1)!]
$\Rightarrow (\text{n}-2) (\text{n}+1)\text{n}= 5\times 4\times3 $ [By comparing] $\therefore \text{n}= 3$ Hence, n = 3
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Question 293 Marks
How many different numbers of six digits each can be formed from the digits 4, 5, 6, 7, 8, 9 when repetition of digits is not allowed?
Answer
First digit of six-digit numbers can be selected in 6 ways.
Second digit of six-digit numbers can be selected in 5 ways
Third digit of six-digit 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 numbers = 6 × 5 × 4 × 3 × 2 × 1 = 720
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Question 303 Marks
Prove that:$\frac{\text{n!}}{(\text{n-r})!}= \text{n}(\text{n}-1)(\text{n}-2).....(\text{n}(\text{n}-\text{r}))$
Answer
We have, $\text{L.H.S.}=\frac{\text{n!}}{(\text{n-r})!}$ $=\frac{\text{n}(\text{n}-1)(\text{n}-2)(\text{n}-3)...(\text{n}-\text{r}+2)(\text{n}-\text{r}+1)(\text{n}-\text{r})!}{(\text{n}-\text{r})!}$ = n(n - 1)(n - 2)(n - 3).....(n - r + 2)(n - r + 1 )) = n(n - 1)(n - 2)(n - 3).....((n - (r - 2))(n - (r - 1 )) = n(n - 1)(n - 2)(n - 3).....(n - (r - 1))= R.H.S.
$\therefore$ L.H.S. = R.H.S.Hence proved.
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Question 313 Marks
Four letters E, K, S and V, one in each, were purchased from a plastic warehouse. How many ordered pairs of letters, to be used as initials, can be formed from them?
Answer
Total number of letters = 4 $\therefore$ The total number of o rd red paris = Number of arrangements of 4 letters, taken two at a time $= \ ^{4}\text{P}_2$$=\frac{4!}{(4-2)!}$
$=\frac{4!}{2!}$
$=\frac{4\times3\times2!}{2!}$
$=4\times3$
$=12$
Hence, the total number of ordered par is = 12
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Question 323 Marks
If P(n ,5) = 20.P(n, 3), find n.
Answer
We have,P(n ,5) = 20.P(n, 3)
$\Rightarrow \frac{\text{n!}}{(\text{n-5})!}=20\times\frac{\text{n!}}{(\text{n-3)}!}$ $\Rightarrow \frac{1}{(\text{n}-5)}!= \frac{20}{(\text{n}-3)(\text{n}-3 -1)(\text{n}-3-2)!}$ $\Rightarrow \frac{1}{(\text{n}-5)!}= \frac{20}{(\text{n}-3)(\text{n}-4)\text{n}-5)!}$ $\Rightarrow \frac{(\text{n}-3)(\text{n}-4)(\text{n}-5)!}{(\text{n}-5)!}= 20$ $\Rightarrow (\text{n}-3)(\text{n}-4) =20$ $\Rightarrow \text{n}^2-4\text{n}-3\text{n}-8=0$ $\Rightarrow \text{n}^2+7\text{n}-8=0$ $\Rightarrow \text{n}^2-8\text{n}+1\text{n}-8=0$ $\Rightarrow \text{n}(\text{n}-8)+1(\text{n}-8)= 0$ $\Rightarrow (\text{n}-8)(\text{n}+1)= 0$ $\Rightarrow \text{n}-3 = 0$ $\Rightarrow \text{n} = 8 [\because\text{n}\neq-1]$ Hence, $ \text{n} = 8$
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Question 333 Marks
Prove that:
n!(n + 2)! = n! + (n + 1)!
Answer
We have,
L.H.S. = n! + (n + 1)!
= n! + (n + 1)(n + 1 - 1)!
= n! + (1 + n + 1)
= n!(n + 2)
= R.H.S.
$\therefore$ n!(n + 2)! = n! + (n + 1)!
Hence, proved.
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Question 343 Marks
In how many ways can three jobs I, II and III be assigned to three persons A, B and C if one person is assigned only one job and all are capable of doing each job?
Answer
Total number of jobs = 3
$\therefore$ the number of ways to assined these job is to
three persons = 3 × 2 × 1
= 6
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Question 353 Marks
Find the total number of ways in which six $'+'$ and four $'−'$ signs can be arranged in a line such that no two $'−'$ signs occur together.
Answer
Total number of $'+'$ signs $= 6$
Total number of $'-'$ signs $= 4$
six $'+'$ signs can be arranged in a row in $=\frac{6!}{6!}=1$ Way [$\because$ All $'+'$ signs are identical]
Now, we are left with seven places in which four different things can be arranged in $^7P_4$ ways but all the four $'-'$ signs are identical, therefore, four $"-'$ signs can be arranged
$=\frac{^7\text{P}_4}{4!}$
$=\frac{7!}{\frac{(7-4)!}{4!}}= \frac{7!}{3!\times4!}$
$=\frac{7\times6\times5\times4!}{3\times2\times4!}=7\times5=35$
Hence, the required number of ways $= 1 \times 35 = 35.$
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Question 363 Marks
How many words can be formed with the letters of the word 'PARALLEL' so that all L's do not come together?
Answer
There are 8 letters in the word 'PARALLEL' out of which A's and 3 are L's and the rest are all distinct. So, total number of words $=\frac{8!}{ 3!\ 2!}$ $=\frac{8\times7\times6\times5\times4\times3!}{2\times1\times3!}$ $= 8\times7\times6\times5\times2$ $=3360.$ Considering all L's together and treating them as one letter we have 6 letters out of which A repeats 2 times and others are distinct. These 6 letters can be arranged in $\frac{6!}{2!}$ ways.So, the number of words in which all L's come together $= \frac{6!}{2!}$
$=\frac{6\times5\times4\times3\times2!}{2!}$ $= 6\times5\times4\times3\times2$ $=360$ Hence, the number of words in which all L's do not come together = 3360 - 360 = 3000
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Question 373 Marks
How many natural numbers not exceeding 4321 can be formed with the digits 1, 2, 3 and 4, if the digits can repeat?
Answer
The given digits are 1, 2, 3 and 4. These digits can be repeated while forming the numbers. So, number of required four digit natural numbers can be found as follows.
Consider four digit natural numbers whose digit at thousandths place is 1.
Here, hundredths place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)
Similarly, tens place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)
Ones place can be filled in 4 ways. (Using the digits 1 or 2 or 3 or 4)
Number of four digit natural numbers whose digit at thousandths place is 1 = 4 × 4 × 4 = 64
Similarly, number of four digit natural numbers whose digit at thousandths place is 2 = 4 × 4 × 4 = 64
Now, consider four digit natural numbers whose digit at thousandths place is 4:
Here, if the digit at hundredths place is 1, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.
If the digit at hundredths place is 2, then tens place can be filled in 4 ways and ones place can also be filled in 4 ways.
If the digit at hundredths place is 3 and the digit at tens place is 1, then ones place can be filled in 4 ways.
If the digit at hundredths place is 3 and the digit at tens place is 2, then ones place can be filled only in 1 way so that the number formed is not exceeding 4321.
Number of four digit natural numbers not exceeding 4321 and digit at thousandths place is 3 = 4 × 4 + 4 × 4 + 4 + 1 = 37
Thus, required number of four digit natural numbers not exceeding 4321 is 64 + 64 + 64 + 37 = 229.
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Question 383 Marks
How many $3-$digit even number can be made using the digits $1, 2, 3, 4, 5, 6, 7,$ if no digits is repeated?
Answer
In order to find the number of even digits, we fix the unit's digit as an even digit.
Fixing the unit's digit as $2:$
Number of arrangements possible $=\ ^6P_2 = 6 \times 5 = 30$
Similarly, fixing the unit's digit as $4:$
Number of arrangements possible $=\ ^6P_2 = 6 \times 5 = 30$
Fixing the unit's digit as $6:$
Number of arrangements possible $=\ ^6P_2 = 6 \times 5 = 30$
$\therefore$ Number of $3-$digit even numbers that can be formed $= 30 + 30 + 30 = 90$
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Question 393 Marks
In how many ways can a lawn tennis mixed double be made up from seven married couples if no husband and wife play in the same set?
Answer
Let two husbands A, B be selected out of seven males in $=\ ^7c_2$ ways.
Excluding their wives, we have to select two ladies C, D out of remaining 5 wives is $=\ ^5c_2$ ways.
Thus, number of ways of selecting the pl ayers for mixed double is $=\ ^7C_2 \times \ ^5C_2$ 
$= 21 \times 10$
$= 210$
Now, suppose $A$ chooses $C$ as partner $(B$ will automatically go to $D)$ or $A$ chooses $D$ as partner $(B$ will automatically go to $C)$ Thus we have, $4$ other ways for teams.
Required number of ways $= 210 \times 4 = 840$
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Question 403 Marks
How many different arrangements can be made by using all the letters in the word 'MATHEMATICS'. How many of them begin with C? How many of them begin with T?
Answer
There are 11 letters in the word 'MATHEMATICS' out of which 2 are M's, 2 are A's, 2 are T's and the rest are all distinet.So, the requisite number of word $=\frac{11!}{2!\ 2!\ 2!}$
If we fix C in the beginning, then the remaining 10 letters can be arranged in $\frac{10!}{2!\ 2!\ 2!}$
If we fix Tin the beginning, 10! then the remainning 10 letters can be arranged in $\frac{10!}{2!\ 2!}$
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Question 413 Marks
How many odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed?
Answer
Any number less than 1000 may be any of a number from one-digit number, two-digit number and three-digit number.One-digit odd number:
3 possible ways are there. These numbers are 3 or 5 or 7. Two-digit odd number:Tens place can be filled up by 3 ways (using any of the digit among 3, 5 and 7) and then the ones place can be filled in any of the remaining 2 digits.
So, there are 3 × 2 = 6 such 2 digit numbers.
Three-digit odd number: Ignore the presence of zero at ones place for some instance. Hundreds place can be filled up in 3 ways (using any of any of the digit among 3, 5 and 7), then tens place in 3 ways by using remaining 3 digits (after using a digit, there will be three digits) and then the ones place in 2 ways. So, there are a total of 3 × 3 × 2 = 18 numbers of 3 digit numbers which includes both odd and even numbers (ones place digit are zero). In order to get the odd numbers, it is required to ignore the even numbers i.e. numbers ending with zero. To obtain the even 3-digit numbers, ones place can be filled up in 1 way (only 0 to be filled), hundreds place in 3 ways (using any of the digit among 3, 5, 7) and then tens place in 2 ways (using remaining 2 digits after filling up hundreds place). So, there are a total of 1 × 3 × 2 = 6 even 3 digit numbers using the digits 0, 3, 5 and 7 (repetition not allowed). So, number of three-digit odd numbers using the digits 0, 3, 5 and 7 (repetition not allowed) = 18 - 6 = 12. Therefore, odd numbers less than 1000 can be formed by using the digits 0, 3, 5, 7 when repetition of digits is not allowed are 3 + 6 + 12 = 21.
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Question 423 Marks
In how many ways can the letters of the word $\text{'STRANGE'}$ be arranged so that.
  1. The vowels come together?
  2. The vowels never come together? and
  3. The vowels occupy only the odd places?
Answer
There are $7$ letters in the word $\text{'STRANGE',}$ including $2$ vowels $(A, E)$ and $5$ consonants $\text{(S, T, R, N, G).}$
  1. Considering $2$ vowels as one letter, we have 6 letters which can be arranged in $^6p_6 = 6!$ ways $A, E$ can be put together in $2!$ ways.
Hence, required number of words
$= 6! \times 2!$
$= 6 \times 5 \times 4 \times 3 \times 2 \times 1 \times 2 \times 1$
$= 720 \times 2$
$= 1440$
  1. The total number of words farmed by using all the letters of the words $\text{'STRANGE'}$ is $ ^7P_7 = 7!$
$= 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$
$= 5040$
So, the total number of words in which vowels are never together
$=$ Total number of words $-$ number of words in which vowels are always together
$= 5040 - 1440$
$= 3600$
  1. There are $7$ letters in the word' $\text{STRANGE'}.$ out of these letters $'A\ '$ and $' E\ '$ are the vowels.
There are $4$ odd places in the word $\text{'STRANGE'.}$ The two vowels can be arranged in $^4p_2$ ways.
The remaining $5$ consonants can be arranged among themselves in $^5p_5$​​​​​​​ ways.
The total number of arrangements.
$=\ ^4\text{P}_2\times\ ^5\text{P}_5$
$=\frac{4!}{2!}\times 5!$
$=1440$
 
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Question 433 Marks
All the letters of the word 'EAMCOT' are arranged in different possible ways. Find the number of arrangements in which no two vowels are adjacent to each other:
Answer
There are $6$ letters in the word 'EAMCOT'. Out of these letters $'E','A'$ and $'O'$ are the three vowels.
The remaining three consonants can be arranged in $^3P_3$ ways. In each of these arrangements $4$ places are created, shown by the cross marks.
X
V
X
V
X
V
X
Since no two vowels are to be placed adjacent to each other, so we may arrange 3 vowels in 4 places in ^$^4P_3$  ways.
The total number of arrangements
$=\ ^3P_3 \times \ ^4P_3$
$= 3! \times 4!$
$= 144$
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Question 443 Marks
There are three copies each of 4 different books. In how many ways can they be arranged in a shelf?
Answer
There are three copies each of 4 different books.
$\therefore$ Total number of copies = 12
$\therefore$ The number of ways in which these copies arranged in a shelf
$=\frac{12!}{3!\ 3!\ 3!\ 3!}$
$=\frac{12!}{(3!)^4}$
Hence, required number of ways
$=\frac{12!}{(3!)^4}$
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Question 453 Marks
How many words can be formed by arranging the letters of the word 'MUMBAI' so that all M's come together?
Answer
There are 6 letters in the word 'MUMBAI' out of which 2 are M's and the rest are all distinct.
Considering both M's together and treating as one letter we have 5 letters. These 5 letters can be arranged in 5! ways.
Hence, the total number of arrangement = 5!
= 5 × 4 × 3 × 2 × 1
= 120
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Question 463 Marks
Find the number of ways in which one can post $5$ letters in $7$ letter boxes.
Answer
Total numbers of letters $= 5$
Total number of letters boxes $= 7$
$\therefore$ The number ways in which one can post $5$ letters in $7$ letter boxes.
$= 7 \times 7 \times 7 \times 7 \times 7 = 7^5$
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Question 473 Marks
prove that:
$\frac{\text{n}!}{\text{(n-r)!r!}}+\frac{\text{n!}}{\text{(n-r+1)!}\text{(r-1)!}}= \frac{\text{(n+1)!}}{\text{r(n-r+1)!}}$
Answer
We have,
$\text{L.H.S.}=\frac{\text{n}!}{\text{(n-r)!r!}}+\frac{\text{n!}}{\text{(n-r+1)!}\text{(r-1)!}}$
$=\frac{\text{n!}}{(\text{n-r})!\text{r}\times\big[\text{(r-1)}!\big]}+\frac{\text{n!}}{\text{(n-r+1)}\big[(\text{n-r})!\big](\text{r-1})!}$
$=\frac{\text{n!}}{\text{(n - r)! }\times{\text{(r-1)!}}}\bigg[\frac{1}{\text{r}}+\frac{1}{\text{n-r+1}}\bigg]$
$=\frac{\text{n!}}{\text{(n - r)! }\times{\text{(r-1)!}}}\bigg[\frac{\text{n-r+1+r}}{\text{r(n-r+1)}}\bigg]$
$=\frac{\text{n!}}{\text{(n - r)! }\times{\text{(r-1)!}}}\bigg[\frac{\text{n+1}}{\text{r(n-r+1)}}\bigg]$
$=\frac{\text{(n+1)}\times\text{n!}}{\text{(n-r+1)!}\times\text{(n-r)}! \times\text{r}\times(\text{r-1})!}$
$=\frac{(\text{n+1}!)}{\text{(n-r+1)!}\times\text{r}!}$
$=\frac{\text{(n+1)!}}{\text{r!}\text{(n-r+1)}!}$
$\text{R.H.S}$
$\therefore$ L.H.S.= R.H.S.
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Question 483 Marks
In how many ways can the letters of the word ASSASSINATION be arranged so that all the S's are together?
Answer
There are 13 letters in the word 'ASSASSINATION' out of which 3 are A's, 4 are S's, 2 are l's, 2 are N's and the rest are all distinct.
Considering all S's together and treating them as one letter we have 10 letters.
These 10 letters can be arranged in $=\frac{10!}{3!\ 2!\ 2!}$
$=\frac{10\times 9\times 8\times7\times6\times5\times4\times 3 !}{3!\times2\times2}$
$=10\times 9\times8\times7\times6\times5$
$=151200$
Hence, the total words are 151200.
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Question 493 Marks
How many three-digit numbers are there, with distinct digits, with each digit odd?
Answer
The odd number digits are 1, 3, 5, 6, 9 Total number of odd digits = 5 $\therefore$ Required number of 3 digit num bars = number of arrangenments of 5 digits by taking 3 at a time $=\ ^{5}\text{P}_3$$=\frac{5!}{(5-3)!}$
$=\frac{5!}{2!}$
$=\frac{5\times4\times3\times2!}{2!}$
$=60$
Hence, total number of 3 digit numbers are 60.
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