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Looking for Class 11 Maths Exercise 7.4 solutions? You’re in the right place! This section provides detailed and easy-to-understand solutions to all the questions from Exercise 7.4 of Chapter 7 – Permutations and Combinations. These solutions are based on the latest NCERT syllabus and cover advanced applications of both permutations and combinations, including mixed problems and real-life scenarios. Whether you’re referring to the Class 11 Ch 7 Exercise 7.4 solutions, solving problems from the NCERT Exemplar Class 11 Maths, or revising the complete topic of Permutation and Combination Class 11, these solutions will help you sharpen your analytical thinking. Download or explore the NCERT Solutions for Class 11 Maths Chapter 7 and take your preparation to the next level!
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Exercise 7.4
1. If \( { }^{\mathrm{n}} \mathrm{C}_{8}={ }^{\mathrm{n}} \mathrm{C}_{2} \), find \( { }^{\mathrm{n}} \mathrm{C}_{2} \).
Answer
It is known that, \( n C_{\mathrm{a}}=n C_{\mathrm{b}}=\mathrm{a}=\mathrm{b} \) or \( \mathrm{m}=\mathrm{a}+\mathrm{b} \)
Therefore,
\({ }^{n} C_{8}={ }^{n} C_{2}=n=8+2=10\)
\(\therefore n_{C_{8}}=10_{C_{2}}=\frac{10!}{2!(10-2)!}=\frac{10!}{2!8!}=\frac{10 \times 9 \times 8!}{2 \times 1 \times 8!}=45\)
2.
(i) Determine n if \( { }^{2 n} C_{2}:{ }^{n} C_{3}=12: 1 \)
Answer
\( \frac{2 n_{c_{3}}}{n_{c_{3}}}=\frac{12}{1}\)
\(=\frac{(2 n)!}{3!(2 n-3)!} \times \frac{3!(n-3)!}{n!}=\frac{12}{1}\)
\(=\frac{2(2 n-1)(2 n-2)}{(n-1)(n-2)}=12\)
\(=\frac{4(2 n-1)(n-1)}{(n-1)(n-2)}=12\)
\(=\frac{(2 n-1)}{(n-2)}=3\)
\(=2 \mathrm{n}-1=3(\mathrm{n}-2)\)
\(=2 n-1=3 n-6\)
\(=3 n-2 n=-1+6\)
\(=\mathrm{n}=5\)
(ii) Determine n if \( { }^{2 n} C_{3}:{ }^{n} C_{3}=11: 1 \)
Answer
\( \frac{2 n_{c_{3}}}{3!(2 n-3)!}=\frac{11}{1} \)
\(=\frac{(2 n)(2 n-1)(2 n-2)(2 n-3)}{(2 n-3)!} \times \frac{(n-3)!}{n(n-1)(n-2)(n-3)!}\)
\(=\frac{2(2 n-1)(2 n-2)}{(n-1)(n-2)}=11\)
\(=\frac{4(2 n-1)(n-1)}{(n-1)(n-2)}=11\)
\(=\frac{4(2 n-1)}{n-2}=11\)
\(=4(2 n-1)=11(\mathrm{n}-2)\)
\(=8 \mathrm{n}-4=11 \mathrm{n}-22\)
\(=11 \mathrm{n}-8 \mathrm{n}=-4+22\)
\(=3 \mathrm{n}=18\)
\(=\mathrm{n}=6\)
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3. How many chords can be drawn through 21 points on a circle?
Answer
For drawing one chord a circle, only 2 points are required.
To know the number of chord that can be drawn through the given 21 points on a circle, the number of combinations have to be counted.
Therefore, there will be as many chord as there are combinations of 21 points taken 2 at a time.
Thus, required number of chord \( ={ }^{23} \mathrm{C}_{2}=\frac{2!}{2!19!}=\frac{21 \times 20}{2}=210 \)
4. In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Answer
A team of 3 boys and 3 girls is to be selected from 5 boys and 4 girls.
3 boys can be selected from 5 boys in \( { }^{5} \mathrm{C}_{3} \) ways.
3 girls can be selected from 4 girls in \( { }^{4} \mathrm{C}_{3} \) ways.
Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected \( ={ }^{5} \mathrm{C}_{3} \times{ }^{4} \mathrm{C}_{3}=\frac{5!}{3!2!} \times \frac{4!}{3!1!} \)
\(=\frac{5 \times 4 \times 3!}{3!\times 2} \times \frac{4 \times 3!}{3!}=10 \times 4=40\)
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5. Find the number of ways of selecting 9 balls from 6 red balls, 5 white balls and 5 blue balls if each selection consists of 3 balls of each color.
Answer
There are a total of 6 red balls, 5 white balls, and 4 blue balls.
9 balls have to be selected in such a way that each selected consists of 3 balls of each colors.
Here,
3 balls can be selected from 6 red balls in \( { }^{6} \mathrm{C}_{3} \) ways.
3 balls can be selected from 5 white balls in \( { }^{5} \mathrm{C}_{3} \) ways.
3 balls can be selected from 5 blue in \( { }^{5} \mathrm{C}_{3} \) ways.
Thus, by multiplication principle, required number of ways of selecting 9 balls.
\(={ }^{6} \mathrm{C}_{3} \times{ }^{5} \mathrm{C}_{3} \times{ }^{5} \mathrm{C}_{3}=\frac{6!}{3!3!} \times \frac{5!}{3!2!} \times \frac{5!}{3!2!}\)
\(=\frac{6 \times 5 \times 4 \times 3!}{3!\times 3 \times 2} \times \frac{5 \times 4 \times 3!}{3!\times 2 \times 1} \times \frac{5 \times 4 \times 3!}{3!\times 2 \times 1}\)
\(=20 \times 10 \times 10=2000\)
6. Determine the number of 5 card combinations out of a deck of 52 cards if there is exactly one ace in each combination.
Answer
In a deck of 52 cards, there are 4 aces, a combination of 5 cards have to be made in which there is exactly one ace.
Then, one ace can be selected in \( { }^{4} \mathrm{C}_{3} \) ways and the remaining 4 cards can be selected out of the 48 cards in \( { }^{48} \mathrm{C}_{4} \) ways.
Thus, by multiplication principle, required number of 5 cards combination
\(={ }^{48} \mathrm{C}_{4} \times{ }^{4} \mathrm{C}_{1}=\frac{48!}{4!44!} \times \frac{4!}{1!3!}\)
\(=\frac{48 \times 47 \times 46 \times 45}{4!\times 3 \times 2 \times 1} \times 4!\)
\(=778320\)
7. In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers?
Answer
Out of 17 players, 5 players are bowlers.
A cricket team of 11 players is to be selected in such a way that there are exactly 4 bowlers.
4 bowlers can be selected in \( { }^{5} \mathrm{C}_{4} \) ways and the remaining 7 players can be selected out of the 12 players in \( { }^{12} \mathrm{C}_{7} \) ways.
Thus, by multiplication principle, required of ways of selecting cricket team
\(={ }^{5} \mathrm{C}_{4} \times{ }^{12} \mathrm{C}_{7}=\frac{5!}{4!1!} \times \frac{12!}{7!5!}=5!\times \frac{12 \times 11 \times 10 \times 9 \times 8}{5!\times 4 \times 3 \times 2 \times 1}=3960\)
8. A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected.
Answer
There are 5 black and 6 red balls in the bag.
2 black balls can be selected out of 5 black balls in\( { }^{5} \mathrm{C}_{2}\) ways and 3 red balls can be selected out of 6 red balls in \({ }^{6} \mathrm{C}_3\) ways.
Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls
\(={ }^{5} \mathrm{C}_{2} \times{ }^{6} \mathrm{C}_3=\frac{5!}{2!3!} \times \frac{6!}{3!3!}=\frac{5 \times 4}{2} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1}=10 \times 20=200\)
9. In how many ways can a student choose a programmer, of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student?
Answer
There are 9 courses available out of which, 2 specific courses are compulsory for every student.
Therefore, every student has to choose 3 courses out of the remaining 7 courses. This can be chosen in \( { }^{7} \mathrm{C}_{3} \) ways.
Thus, required number of ways of choosing the programmer
\(={ }^{7} \mathrm{C}_{3}=\frac{7!}{3!4!}=\frac{7 \times 6 \times 5 \times 4!}{3 \times 2 \times 1 \times 4!}=35\)
