Exercise 6.3 (Revised) - Chapter 7 - Permutations & Combinations - Ncert Solutions class 11 - Maths

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Chapter 6: Permutations & Combinations - NCERT Solutions Class 11 Math

Ex 6.3 Question 1.

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digits is repeated?

Answer.
Given digits from 1 to 9 and repitition of digits are not allowed. We need to a three digit number.
9 different digits taken 3 at a time $={ }^9 P_3=\frac{9!}{3!\times 6!}$
$
=\frac{9 \times 8 \times 7 \times 6!}{6!}=9 \times 8 \times 7=504 \text {. }
$
Ex 6.3 Question 2.

How many 4-digit numbers are there with no digit repeated?

Answer.

Thousandth place can be arranged any of the digits from 1 to 9 because 0 cannot be included. So the number of ways to arrange thousandth place is 9 .

The hundred, Tenths and Unit places can be arranged by any of the digits from 0 to 9 but digits cannot be repeated.

Thus the hundred,tenths and unit places can be arranged in 9 ways.
$\therefore$ there will be as many such 3 digit numbers as there are permutations 9 different digits taken 3 at a time.
$\therefore$ Number of such three digits $=>{ }^9 P_3=\frac{9!}{(9-3)!}=\frac{9!}{6!}=\frac{9 \times 8 \times 7 \times 6!}{6!}=9 \times 8 \times 7=504$
Thus by multiplication principle,

Total number of 4 digit numbers $=9 \times 504=4536$.
Ex 6.3 Question 3.

How many 3-digit even numbers can be made using the digits 1, 2, 3, 4, 6, 7 if no digit is repeated?

Ans. Total number of digits $=6$ and unit place can be filled with any one of the digits 2, 4, 6 .
$\therefore$ Number of ways to arrange unit place $={ }^3 P_1=\frac{3!}{2!}=\frac{3 \times 2 \times 1}{2 \times 1}=3$
Now, the tens and hundreds places can be filled by remaining 5 digits.
$\therefore$ Number of ways to arrange the hundredth and tenth places $={ }^5 P_2=\frac{5!}{3!}=\frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1}=20$.

Therefore, total number of 3 - digit even numbers $=3 \times 20=60$.
Ex 6.3 Question 4.

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.

Even numbers always end with a digit of $0,2,4,6$ or 8
Given digits are 1, 2, 3, 4, 5, Here even numbers end with 2 and 4 .
The number of ways in which unit place can be arranged is $\mathbf{2}$
Since repetition of digits are not allowed, unit place is already arranged with one of the even numbers. So remaining places can be arranged with 4 digits.

Therefore, the number of ways to arrange remainig places is ${ }^4 P_3=\frac{4!}{(4-3)!}=\frac{4!}{1!}=4!=24$

Therefore, the required number of even numbers $=2 \times 24=48$.
Ex 6.3 Question 5.

From a committee of 8 persons, in how many ways can we choose a chairman and a vice chairman assuming one person cannot hold more than one position?

Answer.

We need to form a committee of 8 people with a chairman and a vice - chairman. It is evident that no committee contains two or more chairman and two or more vice - chairman.

So repetition is not allowed.
Therefore, the number of ways to choose a chairman and a vice - chairman is the permutation of 8 things taken 2 at a time.

Required number of ways $={ }^8 P_2=\frac{8!}{(8-2)!}=\frac{8!}{6!}=\frac{8 \times 7 \times 6!}{6!}=8 \times 7=56$.
Ex 6.3 Question 6.

Find $n$ if ${ }^{(n-1)} P_3:{ }^n P_4=1: 9$

Answer.

Given: ${ }^{(n-1)} P_3:{ }^n P_4=1: 9$
$
\begin{aligned}
& \Rightarrow \frac{{ }^{(n-1)} P_3}{{ }^n P_4}=\frac{1}{9} \Rightarrow \frac{(n-1)!}{(n-4)!} \times \frac{(n-4)!}{n!}=\frac{1}{9} \\
& \Rightarrow \frac{(n-1)!}{n!}=\frac{1}{9} \Rightarrow \frac{(n-1)!}{n(n-1)!}=\frac{1}{9} \\
& \Rightarrow \frac{1}{n}=\frac{1}{9} \Rightarrow n=9 .
\end{aligned}
$
Ex 6.3 Question 7.

Find $\gamma$ if:
(i) ${ }^5 \mathrm{P}_r=2{ }^6 \mathrm{P}_{r-1}$
(ii) ${ }^5 \mathrm{P}_r={ }^6 \mathrm{P}_{r-1}$

Answer.

(i) Given: ${ }^5 \mathrm{P}_r=2{ }^6 \mathrm{P}_{r-1}$
$
\begin{aligned}
& \Rightarrow \frac{5!}{(5-r)!}=2 \times \frac{6!}{(7-r)!} \\
& \Rightarrow \frac{5!}{(5-r)!}=\frac{2 \times 6 \times 5!}{(7-r)(6-r)(5-r)!} \\
& \Rightarrow 1=\frac{2 \times 6}{(7-r)(6-r)} \\
& \Rightarrow(7-r)(6-r)=12 \\
& \Rightarrow 42-7 r-6 r+r^2=12 \\
& \Rightarrow r^2-13 r+30=0
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow r^2-3 r-10 r+30=0 \\
& \Rightarrow(r-3)(r-10)=0
\end{aligned}
$
$
\Rightarrow r=3,10
$
$r=10$,is not possible because $r>n$, therefore, $r=3$
(ii) Given: ${ }^5 \mathrm{P}_r={ }^6 \mathrm{P}_{r-1}$
$
\begin{aligned}
& \Rightarrow \frac{5!}{(5-r)!}=\frac{6!}{(7-r)!} \\
& \Rightarrow \frac{5!}{(5-r)!}=\frac{6 \times 5!}{(7-r)(6-r)(5-r)!} \\
& \Rightarrow 1=\frac{6}{(7-r)(6-r)} \\
& \Rightarrow(7-r)(6-r)=6 \\
& \Rightarrow 42-7 r-6 r+r^2=6 \\
& \Rightarrow r^2-13 r+36=0 \\
& \Rightarrow r^2-9 r-4 r+36=0 \\
& \Rightarrow(r-9)(r-4)=0
\end{aligned}
$

$
\Rightarrow r=9 \text { or } r=4
$
$r=9$ is not possible because $r>n$, therefore, $r=4$.
8. How many words, with or without meaning can be formed using all the letters of the word EQUATION, using each letter exactly once?

Answer.

Total number of letters in word " EQUATION " = 8
Number of letters to arranged $=8$ (All distinct)
So it is a permutation of 8 distinct letters taken 8 at a time.

$
\begin{aligned}
& \therefore \text { Number of permutations }={ }^8 P_8=\frac{8!}{(8-8)!}=\frac{8!}{0!}=8! \\
& =8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1=40320
\end{aligned}
$
Ex 6.3 Question 9.

How many words, with or without meaning can be formed using all the letters of the word MONDAY, assuming that no letter is repeated if:
(i) 4 letters are used at a time
(ii) all letters are used at a time
(iii) all letters are used by first letter is a vowel?

Answer.

(ii) We can arrange 6 - letter words from the letters on " MONDAY " without repetition.
It is the permuation of 6 distinct letters taken 6 at a time.
Therefore, required number of words $={ }^6 P_6=\frac{6!}{(6-6)!}=\frac{6!}{0!}=6!$
$
=6 \times 5 \times 4 \times 3 \times 2 \times 1=720 \text {. }
$
(i) Total number of letters in word " MONDAY " $=6$

Number of letters to arranged $=4$ (All distinct)
So it is a permutation of 6 distinct letters taken 4 at a time.
$\therefore$ Number of permutations $={ }^6 P_4=\frac{6!}{(6-2)!}=\frac{6!}{2!}$

$
=6 \times 5 \times 4 \times 3=360
$
(iii) Total number of letters in word MONDAY $=6$, four consonants M, N, D, Y and two vowels $\mathrm{A}$ and $\mathrm{O}$.

There are two different ways to arrange the first letter. This can be done in ${ }^2 \mathrm{P}_1$ ways. Since the letters cannot be repeated and the first letter is already arranged with one of the vowels, so the remaining letters can be arranged with 5 distinct letters. This can be done in ${ }^5 \mathrm{P}_5$ ways.

Therefore total number of words formed $={ }^5 \mathrm{P}_5 \times{ }^2 \mathrm{P}_1=5!\times 2!$

$
=5 \times 4 \times 3 \times 2 \times 1 \times 2 \times 1=240 \text {. }
$
Ex 6.3 Question 10.

In how many of the distinct permutations of the letters in MISSISSIPPI do the four I's not come together?

Answer.

In the word " MISSISSIPPI " ,letter " I " appears 4 times, letter " S " appears 4 times, letter " P " appears 2 times, letter " M " appears once.

Therefore, the number of distinct permutations of the given word "MISSISSIPPI " is
$
\begin{aligned}
& =\frac{11!}{4!\times 4!\times 2!}=\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4!}{4!\times 4 \times 3 \times 2 \times 1 \times 2 \times 1} \\
& =\frac{11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1 \times 2 \times 1}=34650 .
\end{aligned}
$

Now, there are 4 I's in the word " MISSISSIPPI " . When they occur together, treat them as a single letter. This single letter and the remaining 7 letters together we get total 8 letters.

In this 8 letters there are $4^{\prime \prime} \mathrm{S}$ " and 2 " $\mathrm{P}$ " , So it can be arranged in $\frac{8!}{4!\times 2!}=\frac{8 \times 7 \times 6 \times 5 \times 4!}{4!\times 2 \times 1}=840$

Number of distinct permutations $=34650-840=33810$

Ex 6.3 Question 11.

In how many ways the letters of word PERMUTATIONS be arranged if the
(i) words start with $P$ and end with $S$.
(ii) vowels are together.
(iii) there are always 4 letters between $P$ and $S$

Answer.

(i) In PERMUTATIONS number of letters $=12$ and there are 2 T's in it.
No. of ways letters of PERMUTATIONS be arranged so that words always start from $\mathrm{P}$ and end with $S=\frac{10!}{2!}=1814400$
(ii) There are 5 vowels in the word PERMUTATIONS and if they are together are considered them as 1 , so no. of letters then is 8

No. of ways vowels are together $=\frac{8!}{2!} \times 5!=\frac{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2!\times 5 \times 4 \times 3 \times 2 \times 1}{2!}=2419200$

(iii) For 4 letters between $\mathrm{P}$ and $\mathrm{S}, \mathrm{P}$ and $\mathrm{S}$ can be arranged in ways as shown:

As $\mathrm{P}$ can be filled in places $1,2,3,4,5,6$ and 7 , consequently $\mathrm{S}$ can be filled in places $6,7,8,9,10,11$ and 12 leaving 4 places between them. So $\mathrm{P}$ and $\mathrm{S}$ or $\mathrm{S}$ and $\mathrm{P}$ can be filled in $7+7=14$ ways. Now the remaining 10 places can be filled in $\frac{10!}{2!}=1814400$

Therefore No. of ways in which 4 letters between $\mathrm{P}$ and $\mathrm{S}=1814400 \times 14=2540100$