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Permutation(Part-1)| S N Dey| Class 11

In this article, we will discuss the solutions of Ex-7A of the chapter Permutation & Combination of Chhaya mathematics for Class 11, written by S N Dey. As of now, we will first discuss the solutions of Very Short Answer Type Questions. You can definitely follow S N Dey mathematics books Class 11 for better preparation for your board exams.

Very Short Answer Type Questions for Permutation (Ex-7A) | S N dey mathematics | Class 11

$1(i)~~$ If $~~{}^{n+1}P_3=10 \cdot {}^{n-1}P_2,~~$ find $\,\,n.$

Solution.

${}^{n+1}P_3=10 \times {}^{n-1}P_2 \\ \text{or,} ~~ \frac{(n+1)!}{(n+1-3)!}=10 \times \frac{(n-1)!}{(n-1-2)!} \\ \text{or,} ~~\frac{(n+1)!}{(n-2)!}=10 \times \frac{(n-1)!}{(n-3)!} \\ \text{or,} ~~\frac{(n+1)n(n-1)!}{(n-2)(n-3)!}=10 \times \frac{(n-1)!}{(n-3)!} \\ \text{or,} ~~\frac{n(n+1)}{(n-2)}=10 \times 1 \\ \text{or,} ~~ n(n+1)=10(n-2) \\ \text{or,} ~~ n^2+n-10n+20=0 \\ \text{or,} ~~ n^2-9n+20=0 \\ \text{or,} ~~n^2-5n-4n+20=0 \\ \text{or,} ~~ n(n-5)-4(n-5)=0 \\ \text{or,} ~~ (n-5)(n-4)=0 \\ \text{or,} ~~(n-5)=0 ,~~(n-4)=0 \\ \therefore~~n=5,4~~\text{(ans.)}$

$(ii)\quad {}^nP_5= 20. {}^nP_3$

Solution.

${}^nP_5= 20 \cdot {}^nP_3 \\ \text{or,}~~ \frac{n!}{(n-5)!}= 20 \times \frac{n!}{(n-3)!} \\ \text{or,}~~(n-3)!= 20 \times (n-5)! \\ \text{or,}~~(n-3)(n-4)(n-5)!= 20 \times (n-5)! \\ \text{or,}~~(n-3)(n-4)=20 \\ \text{or,}~~ n^2-3n-4n+12-20=0 \\ \text{or,}~~n^2-7n -8=0 \\ \text{or,}~~ n^2-8n+n-8=0 \\ \text{or,}~~n(n-8)+1(n-8)=0 \\ \text{or,}~~ (n-8)(n+1)=0 \\ \text{or,}~~ n-8=0 \qquad [ \because~n \neq -1] \\ \therefore~ n=8~~\text{(ans.)}$

$(iii)~~ {}^{n+1}P_{4} : {}^{n-1}P_3=72 : 5$

Solution.

${}^{n+1}P_{4} : {}^{n-1}P_3=72 : 5 \\ \text{or,}~~ 5 \times {}^{n+1}P_{4} =72 \times {}^{n-1}P_3 \\ \text{or,}~~5 \times \frac{(n+1)!}{(n+1-4)!} = 72 \times \frac{(n-1)!}{(n-1-3)!} \\ \text{or,}~~ 5 \times \frac{(n+1)n(n-1)!}{(n-3)(n-4)!}=72 \times \frac{(n-1)!}{(n-4)!} \\ \text{or,}~~ 5n(n+1)=72(n-3) \\ \text{or,}~~ 5n^2+5n-72n+216=0 \\ \text{or,}~~ 5n^2-67n+216=0 \\ \text{or,}~~ 5n^2-40n-27n+216=0 \\ \text{or,}~~ 5n(n-8)-27(n-8)=0 \\ \text{or,}~~ (5n-27)(n-8)=0 \\ \text{or,}~~ (n-8)=0\qquad [\because ~5n-27 \neq 0] \\ \text{or,}~~ n=8~~\text{(ans.)}$

$(iv)\quad 16. {}^{15}P_n= 13 \cdot {}^{16}P_n$

Solution.

${}^{15}P_n= 13. {}^{16}P_n \\ \text{or,}~~ 16 \cdot \frac{15!}{(15-n)!}=13\cdot\frac{16!}{(16-n)!} \\ \text{or,}~~ \frac{16!}{(15-n)!}=13\cdot\frac{16!}{(16-n)(15-n)!} \\ \text{or,}~~ 1=13\cdot \frac{1}{(16-n)} \\ \text{or,}~~ 16-n=13 \\ \text{or,}~~ 16-13=n \\ \text{or,}~~ n=3~~\text{(ans.)}$

$2.~~$ Prove that, $\quad{}^{2n}P_n={1.3.5…….(2n-1)}2^n$

Solution.

$\text{L.H.S.} \\={}^{2n}P_n\\=\frac{(2n)!}{(2n-n)!} \\= \frac{2n(2n-1)(2n-2)(2n-3)(2n-4) \cdots (2n-n)(2n-\overline{n+1}) \cdots 6.5.4.3.2.1}{n!} \\= \frac{[2n(2n-2)(2n-4)\cdots 6.4.2][(2n-1)(2n-3)(2n-5) \cdots7.5.3.1]}{n!} \\= \frac{2^n[n(n-1)(n-2)…3.2.1][(2n-1)(2n-3)(2n-5) \\ \cdots 7.5.3.1]}{n!} \\= \frac{2^n.n![(2n-1)(2n-3)(2n-5)\cdots 7.5.3.1]}{n!} \\=\{1.3.5\cdots(2n-1)\}2^n\\ =\text{R.H.S}$

$3.~~$ If $~~{}^9P_5 +5 \cdot{}^9P_4={}^{10}P_r ,~~$ find $~~r.$

Solution.

$\quad {}^9P_5 +5 \cdot{}^9P_4={}^{10}P_r \\ \text{or,}~~\frac{9!}{(9-5)!}+5.\frac{9!}{(9-4)!}=\frac{10!}{(10-r)!} \\ \text{or,}~~ \frac{9!}{4!}+5.\frac{9!}{5!}=\frac{10!}{(10-r)!} \\ \text{or,}~~\frac{9!}{4!}+5.\frac{9!}{5.4!}=\frac{10!}{(10-r)!} \\ \text{or,}~~ \frac{9!}{4!}+\frac{9!}{4!}=\frac{10.9!}{(10-r)!} \\ \text{or,}~~ 2.\frac{9!}{4!}=\frac{10.9!}{(10-r)!} \\ \text{or,}~~ \frac{2}{4!}=\frac{10}{(10-r)!} \\ \text{or,}~~\frac{1}{4!}=\frac{5}{(10-r)!} \\ \text{or,}~~ 5.4!=(10-r)! \\ \text{or,}~~ 5!=(10-r)! \\ \text{or,}~~ 5=10-r \\ \text{or,}~~ r=10-5 =5~~\text{(ans.)}$

$4.~~$ In how many different ways can $~4~$ prizes be distributed among $~10~$ students? (Each students being eligible for one prize.)

Solution.

The first prize can be given in $~10~$ different ways among $~10~$ students. Once the first prize has been awarded, the second prize can be awarded among nine students in $~9~$ different ways. Similarly, the third prize can be awarded among $~8~$ students in $~8~$ different ways and the last prize can be given in seven ways among the remaining $~7~$ students.

So, the total number ways in which the $~4~$ prizes can be distributed among $~10~$ students is given by $~10 \times 9 \times 8 \times 7 = 5040.$

Alternatively, the number of different ways in which $~4~$ prizes can be distributed among $~10~$ students is given by :

${}^{10}P_4 \\ =\frac{n!}{(n-k)!} \\ =\frac{10!}{(10-4)!}\\=\frac{10!}{6!}\\ =\frac{10.9.8.7.{6!}}{{6!}}\\ =5040 \,\,\,\text{(ans.)}$

$5.~~$ On entering a certain town, 4 travellers find that there are 5 hotels in the town. If no two travellers get into the same hotel, find in how many different ways can the travellers reside in the hotels?

Solution.

Let the four travellers be denoted by $~~a,~b,~c,~d.$

Now, $~a~$ has a choice of $~5~$ hotels, and when he has made his choice in one way, the second traveller $(~b~)$ has a choice of $~5-1=4~$; therefore the first two can make their choice in $~~ 5\times 4~~$ ways and then, $~~c~~$ can select his hotel in $~3~$ ways. Finally, last traveller $(~d~)$ can select his hotel in $~2~$ ways.

So, total number of different ways four travellers can reside in the hotels is given by :

$5\times 4 \times 3 \times 2=120.$

Alternatively, four passengers will stay in five hotels. So if two travellers are not together in the same hotel, the number of arrangement is :

${}^{5}P_4\\=\frac{5!}{(5-4)!}\\=\frac{5!}{1!}\\=\frac{5 \times 4 \times 3 \times 2 \times 1}{1}\\=120$

$6.~~$ 12 ferry-steamers ply between Chandpal Ghat and Botanical Garden. In how many different ways can a person go from Chandpal Ghat to Botanical Garden in one steamer and return in a different steamer?

Solution.

Clearly, you need two steamers to come and go every time. So the number of arrivals and departures by $~12~$ steamers is given by :

${}^{12}P_2\\=\frac{12!}{(12-2)!}\\~=\frac{12!}{10!}\\~=\frac{12\cdot 11 \cdot{10!}}{{10!}}\\~={132}~~\text{(ans.)}$

$7.~~$ There are 12 stations on a certain railway line. How many different kinds of tickets of class II must be printed in order that a passenger may go from any one station to any other ?

Solution.

Clearly, number of printed tickets for going from one station to another out of twelve stations is given by :

${}^{12}P_2\\~=\frac{12!}{(12-2)!}\\~=\frac{12!}{10!}\\~=\frac{12 \cdot11\cdot{10!}}{{10!}}\\~={132}~~\text{(ans.)}$

$8.~~$ How many different permutations can be made by taking all the letters of the word BENGALI ?

Solution.

The word BENGALI has 7 letters and the letters are different. So when the letters of the word BENGALI are arranged all at once, the total number of arrangement is given by :

${}^{7}P_7\\~=\frac{n!}{(n-k)!}\\~=\frac{7!}{(7-7)!}\\~=\frac{7!}{0!}\\~=\frac{7 \cdot 6 \cdot 5\cdot4\cdot 3\cdot 2 \cdot 1}{1}\\~={5040}~~\text{(ans.)}$

$9.~~$ Find the number of ways in which the letters of the word DRAUGHT can be arranged so that the vowels are always together.

Solution.

The word DRAUGHT has a total of seven letters. It has two vowels namely A and U.

If we take these two vowels as one letter, the total number of letters is $~6.$ These 6 letters can be arranged among themselves in $~ {}^{6}P_6=6!~$ ways.

Again the two vowels can be arranged between themselves in $~ {}^{2}P_2=2!~$ ways.

Hence, the number of ways in which the letters of the word DRAUGHT can be arranged so that the vowels are always together is

$=6! \times 2!=720 \times 2=1440~~\text{(ans.)}$

$10.~~$ How many numbers each lying between $~100~$ and $~1000~$ can be formed with the digits $~2, 4, 6, 8, 9~$ each of the digit occurring once in each number?

Solution.

According to the problem, here total number of digits is $~~5.$ The numbers lying between $~100~$ and $~1000~$ are three digits. So, total number of $~3~$ digit numbers formed with the given digits each of the digit occurring once in each number is :

${}^{5}P_3=\frac{5!}{(5-3)!}=\frac{5\cdot 4 \cdot 3 \cdot2!}{2!}=5\cdot 4 \cdot 3=60$

$11.~~$ Find the number of different permutations that can be made out of the letters of the following words:

Solution.

$(i)~~$ COMMERCE $~~(ii)~~$ ACCOUNTANT $~~(iii)~~$ ENGINEERING $~(iv)~~$ STATISTICS $~~(v)~~$ SUCCESS

Solution.

$(i)~~$ Number of letters and the no. of times those letters appear in COMMERCE is shown below.

So, the number of different permutations that can be made out of the letters of ACCOUNTANT is

$=\frac{8!}{2!\times 2! \times 2!}\\~~~~=\frac{8 \times 7!}{8}\\~~~~=7!\\~~~~=5040~~\text{(ans.)}$

For  the rest of the Very Short Answer Type Questions' Solutions, consult the following Book Preview .

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