Punjab State Board PSEB 11th Class Maths Book Solutions Chapter 7 Permutations and Combinations Ex 7.4 Textbook Exercise Questions and Answers.
PSEB Solutions for Class 11 Maths Chapter 7 Permutations and Combinations Ex 7.4
Question 1.
If (\({ }^{n} \mathbf{C}_{8}={ }^{n} C_{2}\)), find (\({ }^{n} \mathbf{C}_{2}\)).
Answer.
It is known that, (\({ }^{n} \mathbf{C}_{a}={ }^{n} C_{2}\))
⇒ a =
⇒ n = a +
Therefore, (\({ }^{n} \mathbf{C}_{8}={ }^{n} C_{2}\))
⇒ n = 8 + 2 = 10
∴ \({ }^{n} \mathbf{C}_{2}={ }^{10} C_{2}\)
= \(\frac{10 !}{2 !(10-2) !}=\frac{10 !}{2 ! 8 !}\)
= \(\frac{10 \times 9 \times 8 !}{2 \times 1 \times 8 !}\) = 45.
Question 2.
Determine n if
(i) (\(\left.{ }^{2 n} C_{3}:{ }^{n} C_{3}\right)\)) = 12 : 1
(ii) \(\left.{ }^{2 n} C_{3}:{ }^{n} C_{3}\right)\)) = 11 : 1
Answer.
(i) Given, (\(\left.{ }^{2 n} C_{3}:{ }^{n} C_{3}\right)\)) = 12 : 1
(ii) \(\left.{ }^{2 n} C_{3}:{ }^{n} C_{3}\right)\)) = 11 : 1
or \(\frac{2 n(2 n-1)(2 n-2)}{1 \times 2 \times 3} \div \frac{n(n-1)(n-2)}{1 \times 2 \times 3}=\frac{11}{1}\)
or \(\frac{4 n(n-1)(2 n-1)}{6} \times \frac{6}{n(n-1)(n-2)}=\frac{11}{1}\)
or 4 (2n – 1) = 11 (n – 2)
or 8n – 4 = 11n – 22
∴ 3n = 18
or n = 6.
Question 3.
How mpny chords can be drawn through 21 points on a circle?
Answer.
For drawing one chord on a circle, only 2 points are required.
To know the number of chords 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 chords as there are combinations of 21 points taken 2 at a time.
Thus, required number of chords = \({ }^{21} C_{2}=\frac{21 !}{2 !(21-2) !}\)
= \(\frac{21 !}{2 ! 19 !}=\frac{21 \times 20}{2}\)
= 210.
Question 4.
In how many ways can a team of 3 boys and 3 girls be selected from 5 boys and 4 girls?
Answer.
There are 5 boys and 4 girls. We have to select 3 out of 5 boys and 3 out of 4 girls.
∴ Number of ways of selection = \({ }^{5} C_{3} \times{ }^{4} C_{3}\)
= \(\frac{5 !}{3 ! 2 !}=\frac{4 !}{3 ! 1 !}\)
= \(\frac{5 \times 4}{2 \times 1} \times \frac{4}{1}\)
= 10 × 4 = 40.
Question 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 colour.
Answer.
There are a total of 6 red balls, 5 white balls, and 5 blue balls.
9 balls have to be selected in such a way that each selection consists of 3 balls of each colour.
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 balls 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 × 10 × 10 = 2000.
Question 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 \({ }^{4} \mathrm{C}_{1}\) 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 card combinations
= \({ }^{48} C_{4} \times{ }^{4} C_{1}\)
= \(\frac{48 !}{4 ! 44 !} \times \frac{4 !}{1 ! 3 !}\)
= \(\frac{48 \times 47 \times 46 \times 45}{4 \times 3 \times 2 \times 1} \times 4\) = 778320.
Question 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 number 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 × \(\frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1}\)
= 3690.
Question 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 bails 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 = \({ }^{5} C_{2} \times{ }^{6} C_{3}\)
= \(\frac{5 !}{2 ! 3 !} \times \frac{6 !}{3 ! 3 !}\)
= \(\frac{5 \times 4}{2} \times \frac{6 \times 5 \times 4}{3 \times 2 \times 1}\)
= 10 × 20 = 200.
Question 9.
In how many ways can a student choose a programme 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 \(\) ways.
Thus, required number of ways of choosing the programme.
= \({ }^{7} \mathrm{C}_{3}=\frac{7 !}{3 ! 4 !}\)
= \(\frac{7 \times 6 \times 5 \times 4 !}{3 \times 2 \times 1 \times 4 !}\) = 35.