Miscellaneous Exercise Class 11 Chapter 7 Permutations and Combinations

In the word DAUGHTER, there are 3 vowels namely, \( \mathrm{A}, \mathrm{U} \), and E and 5 consonants namely, D, G, H, T, and R.
Number of wats of selecting 2 vowels of 3 vowels \( ={ }^{3} \mathrm{C}_{2}=3 \)
Number of ways of selecting 3 consonants out of 5 consonants \( ={ }^{5} \mathrm{C}_{3}= \) 10
Therefore, the number of combination of 2 vowels and 3 consonants \( =3 \) \( \times 10=30 \)
Each of these 30 combinations of 2 vowels and 3 consonants can be arranged among themselves in 5! Ways.
Hence, required numbers of different words \( =30 \times 5!=3600 \)

In the word EQUATION, there are 5 vowels namely, \( A, E, \mathrm{I}, \mathrm{O} \), and U and consonants, namely \( \mathrm{Q}, \mathrm{T} \), and N .
Since all the vowels and consonants have to occur together, both (AEIOU) and (QTN) can be assumed as single objects. Then the permutations of these taken all at a time are counted.
This number would be \( { }^{2} \mathrm{P}_{2}=2 !\)
Corresponding to each of these permutations, there are 5! Permutations of the five vowels taken all at a time and 3! Permutations if the 3 consonants taken all at a time.
Hence, by multiplication principle, required of words \( =2!\times 5!3!= 1440\)

(i) exactly 3 girls
A committee of 7 has to be formed from 9 boys and 4 girls.
Since exactly 3 girls are to be there in every committee must consist of \( (7-3)=4 \) boys only
Thus, in this case, required number of ways \( -{ }^{4} \mathrm{C}_{3} \times{ }^{3} \mathrm{C}_{4}=\frac{4!}{3!2!} \times!\frac{9!}{4!5!}= \) \( \times \frac{9 \times 8 \times 7 \times 6 \times 5!}{4 \times 3 \times 2 \times 5!}=504 \)
(ii) since, at least 3 girls can be there in every committee, the committee can consist of
(a) 3 girls and 4 boys or
(b) \( =4 \) girls and 3 boys
3 girls and 4 can be selected in \( { }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4} \) ways.
4 girls and 3 boys can be selected in \( { }^{4} \mathrm{C}_{4} \times{ }^{9} \mathrm{C}_{3} \) ways.
Therefore, in this case, required number of ways \( ={ }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{4} \times{ }^{9} \mathrm{C}_{3} \) \( =504+84=588 \)
(iii) since at most 3 girls are to be there in every committee, the committee can consist of
(a) 3 girls and 4 boys
(b) 2 girls and 5 boys
(c) 1 girl and 6 boys
(d) No girl and 7 boys
3 girls and 4 boys can be selected in \( { }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4} \) ways.
2 girls and 5 boys can be selected in \( { }^{4} \mathrm{C}_{2} \times{ }^{9} \mathrm{C}_{5} \) ways.
1 girl and 6 boys can be selected in \( { }^{4} \mathrm{C}_{2} \times{ }^{9} \mathrm{C}_{6} \) ways.
No girl and 7 boys can be selected in 4C0 9C7 ways.
Therefore, in this case, required number of ways.
\(={ }^{4} \mathrm{C}_{3} \times{ }^{9} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{2} \times{ }^{9} \mathrm{C}_{5}+{ }^{4} \mathrm{C}_{2} \times{ }^{9} \mathrm{C}_{6}+{ }^{4} \mathrm{C}_{0} \times{ }^{9} \mathrm{C}_{7}\)
\(=\frac{4!}{3!1!} \times \frac{9!}{4!5!}+\frac{4!}{2!2!} \times \frac{9!}{2!2!}+\frac{4!}{1!3!} \times \frac{9!}{6!3!}+\frac{4!}{0!4!} \times \frac{9!}{7!2!}\)
\(=504+756+336+36\)
\(=1632\)

In the given word EXAMINATION, there are 11 letters out of which, A I and N appear 2 times and all the other letters appear only once.
The word that will be listed before the words starting with E in a dictionary will be the words that start with A only
Therefore, to get the numbers of words starting with The Letter A is fixed at the extreme left position and then the remaining 10 letters taken all at a time are rearranged.
Since there are 2 IS and NS in the remaining 10 letters.
Numbers of the words starting with \( A=\frac{10!}{2!2!}=907200 \)
Thus, the required numbers of words are 907200 .

A number is divisible by 10 if its unit digits is 0 .
Therefore, 0 is fixed at the unit place.
Therefore, there will be as many ways as there are ways of filling 5 vacant places \(\begin{array}{|c|c|}\hline & & & & & 0 \\\hline\end{array}\) in succession by the remaining 5 digits (i. e., 1, 3, 5, 7, and 9 ) the 5 vacant places can be filled in \(5 !\) Ways.
Hence, required number of \( 6- \) digit numbers \( =5!=120 \)

2 different vowels and 2 different consonants are to be selected from the English alphabet. Since there are 5 vowels in the English alphabet, numbers of ways of selecting 2 different vowels from the alphabet \(={ }^{5} \mathrm{C}_{2}=\frac{5!}{2!3!}=10\)
Since there are 21 consonants in the English alphabet numbers of ways of selecting 2 different consonants from the alphabet
\(={ }^{21} \mathrm{C}_{2}=\frac{21!}{2!19!}=210\)
Therefore, number of combinations of 2 different vowels and 2 different consonants \( =10 \times 210=2100 \)
Each of these 2100 combinations has 4 letters, which can be arranged among themselves in \(4 !\) Ways.
Therefore, required numbers of words \( =2100 \times 4!=50400 \).

It is given that the question paper consist of 12 question divided into two parts - part I and part II, containing 5 and 7 questions, respectively.
A student has to attempt 8 questions, selecting at least 3 from each part. This can be done as follows.
(a) 3 questions from part I and 5 questions from part II
(b) 4 questions from part I and 4 questions from part II
(c) 5 questions form part I and 3 question from part I
3 questions from part I and 5 questions from part II can be selected in \( { }^{5} \mathrm{C}_{3} \times{ }^{7} \mathrm{C}_{3} \) ways.
4 questions from part I and 4 questions from part II can be selected in \( { }^{5} \mathrm{C}_{4} \times{ }^{7} \mathrm{C}_{4} \) ways.
5 questions from part I and 3 questions from part II can be selected in \( { }^{5} \mathrm{C}_{5} \times{ }^{7} \mathrm{C}_{3} \) ways.
Thus, required number of ways of selecting questions \( =^5 \mathrm{C}_3 \times ^7 \mathrm{C}_5+ ^5 \mathrm{C}_4 \times ^7 \mathrm{C}_4+^5 \mathrm{C}_5 \times ^7 \mathrm{C}_3 \)
\(=\frac{5!}{2!3!} \times \frac{7!}{2!5!}+\frac{5!}{4!1!} \times \frac{4!}{4!3!}+\frac{5!}{5!0!} \times \frac{7!}{3!4!}\)
\(=210+175+35=420\)

From a deck of 52 cards, 5 - card combinations have to be made in such a way that un each selection of 5 cards, there is exactly one king.
In a deck of 52 cards, there are 4 kings.
1 king can be selected out of 4 kings in \( ^4\mathrm{C}_{1} \) ways.
4 cards out of the remaining 48 cards can be selected in \( ^48 _{C_{4}} \) ways.
Thus, the required number of \( 5- \) card combinations is \( { }^{4} \mathrm{C}_{1} \times{ }^{48} \mathrm{C}_{4} \)

4 men and 4 women are to be seated in a row such that the women occuy the even places.
The 5 men can be seated in \(5!\) Ways. For each arrangement the 4 women can be seated only at the cross marked places (so that women occupy the even places).
Therefore, then woman can be seated in \(4 !\) Ways.
Thus, possible number of arrangement \( -4!\times 5!=24 \times 120=2880 \)

There are 2 options
\(\begin{array}{|l|l|}
\hline \text{Either all 3 will go} & \text{None of them will go} \\
\hline \begin{array}{l}
\text{The remaining students in class} \\
\text{Tare: 25-3=22}
\end{array} & \text{The students going will be 10} \\
\hline \begin{array}{l}
\text{Number of students remained to} \\
\text{be chosen for party =7 }
\end{array} & \begin{array}{l}
\text{Remaining students eligible for} \\
\text{going =22 }
\end{array} \\
\hline \begin{array}{l}
\text{Number of ways of choosing the} \\
\text{remaining 22 students }{}^{-22} C_7
\end{array} & \begin{array}{l}
\text{Number of ways in which these }\\
\text{10 students can be selected are}
\end{array} \\
\hline =\frac{22!}{7!15!}=170544 & ^{22} C_10 \\
\hline & =\frac{22!}{10!12!}=646646 \\
\hline
\end{array}\)
Total numbers of ways in which students can be chosen are
\(\begin{array}{l}
=170544+646646 \\
=817190
\end{array}\)

In the given word ASSASSINATION, the letter A appears 3 times, \( S \) appears 4 times, I appear 2 times, N appears 2 times, and all the other letters appears only once. Since all the words have to be arranged in such a way that all the SS are together, SSSS is treated as a single object for the time being. This single objects together with the remaining 9 objects will account for 10 objects.
These 10 objects in which there are 3 As, 2 Is and 2 Ns can be arranged in \( \frac{10!}{3!2!2!} \) ways.
Thus, required numbers of ways of arranging the letters of the given word \( =\frac{10!}{3!2!2!}=151200 \)