Miscellaneous Example (Revised) - Chapter 6 Permutations & Combinations class 11 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

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

Example 1 

How many words, with or without meaning, each of 3 vowels and 2 consonants can be formed from the letters of the word INVOLUTE ?
Solution

In the word INVOLUTE, there are 4 vowels, namely, I,O,E,Uand 4 consonants, namely, N, V, L and T.

The number of ways of selecting 3 vowels out of $4={ }^4 \mathrm{C}_3=4$.
The number of ways of selecting 2 consonants out of $4={ }^4 \mathrm{C}_2=6$.
Therefore, the number of combinations of 3 vowels and 2 consonants is $4 \times 6=24$.

Now, each of these 24 combinations has 5 letters which can be arranged among themselves in 5 ! ways. Therefore, the required number of different words is $24 \times 5!=2880$.

Example 2

A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has (i) no girl ? (ii) at least one boy and one girl ? (iii) at least 3 girls?

Solution

(i) Since, the team will not include any girl, therefore, only boys are to be selected. 5 boys out of 7 boys can be selected in ${ }^7 \mathrm{C}_5$ ways. Therefore, the required number of ways $={ }^7 \mathrm{C}_5=\frac{7!}{5!2!}=\frac{6 \times 7}{2}=21$
(ii) Since, at least one boy and one girl are to be there in every team. Therefore, the team can consist of
(a) 1 boy and 4 girls
(b) 2 boys and 3 girls
(c) 3 boys and 2 girls
(d) 4 boys and 1 girl.
1 boy and 4 girls can be selected in ${ }^7 \mathrm{C}_1 \times{ }^4 \mathrm{C}_4$ ways.
2 boys and 3 girls can be selected in ${ }^7 \mathrm{C}_2 \times{ }^4 \mathrm{C}_3$ ways.
3 boys and 2 girls can be selected in ${ }^7 \mathrm{C}_3 \times{ }^4 \mathrm{C}_2$ ways.
4 boys and 1 girl can be selected in ${ }^7 \mathrm{C}_4 \times{ }^4 \mathrm{C}_1$ ways.
Therefore, the required number of ways
$
\begin{aligned}
& ={ }^7 \mathrm{C}_1 \times{ }^4 \mathrm{C}_4+{ }^7 \mathrm{C}_2 \times{ }^4 \mathrm{C}_3+{ }^7 \mathrm{C}_3 \times{ }^4 \mathrm{C}_2+{ }^7 \mathrm{C}_4 \times{ }^4 \mathrm{C}_1 \\
& =7+84+210+140=441
\end{aligned}
$

(iii) Since, the team has to consist of at least 3 girls, the team can consist of
(a) 3 girls and 2 boys, or
(b) 4 girls and 1 boy.

Note that the team cannot have all 5 girls, because, the group has only 4 girls.
3 girls and 2 boys can be selected in ${ }^4 \mathrm{C}_3 \times{ }^7 \mathrm{C}_2$ ways.
4 girls and 1 boy can be selected in ${ }^4 \mathrm{C}_4 \times{ }^7 \mathrm{C}_1$ ways.

Therefore, the required number of ways
$
={ }^4 \mathrm{C}_3 \times{ }^7 \mathrm{C}_2+{ }^4 \mathrm{C}_4 \times{ }^7 \mathrm{C}_1=84+7=91
$

Example 3

Find the number of words with or without meaning which can be made using all the letters of the word AGAIN. If these words are written as in a dictionary, what will be the $50^{\text {th }}$ word?

Solution

There are 5 letters in the word AGAIN, in which A appears 2 times. Therefore, the required number of words $=\frac{5!}{2!}=60$.

To get the number of words starting with $\mathrm{A}$, we fix the letter $\mathrm{A}$ at the extreme left position, we then rearrange the remaining 4 letters taken all at a time. There will be as many arrangements of these 4 letters taken 4 at a time as there are permutations of 4 different things taken 4 at a time. Hence, the number of words starting with $A=4!=24$. Then, starting with $G$, the number of words $=\frac{4!}{2!}=12$ as after placing $G$ at the extreme left position, we are left with the letters A, A, I and N. Similarly, there are 12 words starting with the next letter I. Total number of words so far obtained $=24+12+12=48$.
The $49^{\text {th }}$ word is NAAGI. The $50^{\text {th }}$ word is NAAIG.

Example 4

How many numbers greater than 1000000 can be formed by using the digits $1,2,0,2,4,2,4$ ?

Solution

Since, 1000000 is a 7-digit number and the number of digits to be used is also 7. Therefore, the numbers to be counted will be 7-digit only. Also, the numbers have to be greater than 1000000 , so they can begin either with 1,2 or 4 .
The number of numbers beginning with $1=\frac{6!}{3!2!}=\frac{4 \times 5 \times 6}{2}=60$, as when 1 is

fixed at the extreme left position, the remaining digits to be rearranged will be $0,2,2,2$, 4,4 , in which there are $3,2 s$ and $2,4 s$.
Total numbers begining with 2
$
=\frac{6!}{2!2!}=\frac{3 \times 4 \times 5 \times 6}{2}=180
$
and total numbers begining with $4=\frac{6!}{3!}=4 \times 5 \times 6=120$

Therefore, the required number of numbers $=60+180+120=360$.
Alternative Method
The number of 7 -digit arrangements, clearly, $\frac{7!}{3!2!}=420$. But, this will include those numbers also, which have 0 at the extreme left position. The number of such arrangements $\frac{6!}{3!2!}$ (by fixing 0 at the extreme left position) $=60$.
Therefore, the required number of numbers $=420-60=360$.

Example 5

In how many ways can 5 girls and 3 boys be seated in a row so that no two boys are together?

Solution

Let us first seat the 5 girls. This can be done in 5 ! ways. For each such arrangement, the three boys can be seated only at the cross marked places.
$
\times \mathrm{G} \times \mathrm{G} \times \mathrm{G} \times \mathrm{G} \times \mathrm{G} \times \text {. }
$

There are 6 cross marked places and the three boys can be seated in ${ }^6 \mathrm{P}_3$ ways. Hence, by multiplication principle, the total number of ways
$
\begin{aligned}
& =5!\times{ }^6 P_3=5!\times \frac{6!}{3!} \\
& =4 \times 5 \times 2 \times 3 \times 4 \times 5 \times 6=14400
\end{aligned}
$