NCERT Solutions for Class 11 Maths Chapter 7 – Permutations and Combinations Ex 7.4
Page No 153:
Question 1:
If
, find
.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5557/Chapter%207_html_680771a9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5557/Chapter%207_html_m1974bb91.gif)
Answer:
It is known that,![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5557/Chapter%207_html_m7207bbe5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5557/Chapter%207_html_m7207bbe5.gif)
Therefore,
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5557/Chapter%207_html_m5594c505.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5557/Chapter%207_html_m6348a58e.gif)
Question 2:
Determine n if
(i)
(ii) ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5558/Chapter%207_html_m7f21859f.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5558/Chapter%207_html_2bfd765e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5558/Chapter%207_html_m7f21859f.gif)
Answer:
(i)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5558/Chapter%207_html_5112b759.gif)
(ii)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5558/Chapter%207_html_m1a797658.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5558/Chapter%207_html_238cfa84.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5558/Chapter%207_html_2d649dc.gif)
Question 3:
How many 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 =![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/4227/chapter%207_html_m559d4c03.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/4227/chapter%207_html_m559d4c03.gif)
Question 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
ways.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5559/Chapter%207_html_m4fd60ef3.gif)
3 girls can be selected from 4 girls in
ways.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5559/Chapter%207_html_2e2a392b.gif)
Therefore, by multiplication principle, number of ways in which a team of 3 boys and 3 girls can be selected ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5559/Chapter%207_html_4fb2c9c9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5559/Chapter%207_html_4fb2c9c9.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5559/Chapter%207_html_429cf5a.gif)
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
ways.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5658/Ch-7_html_6c68f037.gif)
3 balls can be selected from 5 white balls in
ways.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5658/Ch-7_html_m4fd60ef3.gif)
3 balls can be selected from 5 blue balls in
ways.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5658/Ch-7_html_m4fd60ef3.gif)
Thus, by multiplication principle, required number of ways of selecting 9 balls
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5658/Ch-7_html_m12894072.gif)
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
ways and the remaining 4 cards can be selected out of the 48 cards in
ways.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5561/Chapter%207_html_m49b3dcb5.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5561/Chapter%207_html_m7adbb88d.gif)
Thus, by multiplication principle, required number of 5 card combinations ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5561/Chapter%207_html_m3c5cfb5e.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5561/Chapter%207_html_m3c5cfb5e.gif)
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
ways and the remaining 7 players can be selected out of the 12 players in
ways.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5562/Chapter%207_html_m40103821.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5562/Chapter%207_html_20ec8583.gif)
Thus, by multiplication principle, required number of ways of selecting cricket team ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5562/Chapter%207_html_m72a3fbba.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5562/Chapter%207_html_m72a3fbba.gif)
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 balls can be selected out of 5 black balls in
ways and 3 red balls can be selected out of 6 red balls in
ways.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5563/Chapter%207_html_m4147acd7.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5563/Chapter%207_html_6c68f037.gif)
Thus, by multiplication principle, required number of ways of selecting 2 black and 3 red balls ![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5563/Chapter%207_html_1f114cce.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5563/Chapter%207_html_m1b7e4b87.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5563/Chapter%207_html_1f114cce.gif)
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5563/Chapter%207_html_m1b7e4b87.gif)
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.
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5564/Chapter%207_html_m56c5ec0.gif)
Thus, required number of ways of choosing the programme
![](https://img-nm.mnimgs.com/img/study_content/curr/1/11/11/167/5564/Chapter%207_html_m37c020e0.gif)