CBSE Class 11 Mathematics Permutations And Combinations Worksheet Set 07

CBSE Class 11 Mathematics Worksheet - Permutations and Combinations (6). Students can download these worksheets and practice them. This will help them to get better marks in examinations. Also refer to other worksheets for the same chapter and other subjects too. Use them for better understanding of the subjects.

Question. 4 cards out of 52 cards are chosen. Find the number of ways in which:
1. 4 cards are chosen.
2. 4 cards are of the same suit.
3. 4 cards belong to 4 different suits.
4. All are face cards.
5. Two are red & two are black.
6. 4 cards are of the same colour.
Answer:
1. 4 cards are chosen:
4 cards out of 52 cards can be chosen in \( {}^{52}C_4 \) ways:
\( = \frac{52!}{4! 48!} = 270725 \) ans.

2. 4 cards out of same suit:
There are 4 suits (Diamond, Club, Heart, Spade), each having 13 cards.
Number of ways of selecting 4 diamonds out of 13 diamond cards \( = {}^{13}C_4 \).
Similarly, there are \( {}^{13}C_4 \) ways for selecting 4 spades, 4 clubs, and 4 hearts.
\( \therefore \) Required number of ways of selection \( = {}^{13}C_4 + {}^{13}C_4 + {}^{13}C_4 + {}^{13}C_4 = 4 \times {}^{13}C_4 = 2860 \) ans.

3. 4 cards belong to 4 different suits:
We have to select 1 card from each suit.
1 diamond out of 13 diamonds can be selected in \( = {}^{13}C_1 \) ways.
Similarly, 1 club, 1 heart, and 1 spade can each be selected in \( {}^{13}C_1 \) ways.
\( \therefore \) Required number of ways of selection \( = {}^{13}C_1 \times {}^{13}C_1 \times {}^{13}C_1 \times {}^{13}C_1 = 13 \times 13 \times 13 \times 13 = 13^4 = 28561 \) ans.

4. All are face cards:
There are 12 face cards (4J, 4Q, 4K) in a deck. 4 face cards out of 12 can be selected in \( {}^{12}C_4 \) ways:
\( = \frac{12!}{4! 8!} = 495 \) ans.

5. Two are red & two are black:
There are 26 red cards and 26 black cards.
2 red cards out of 26 red cards can be selected in \( = {}^{26}C_2 \) ways.
2 black cards out of 26 black cards can be selected in \( = {}^{26}C_2 \) ways.
\( \therefore \) Required number of selections \( = {}^{26}C_2 \times {}^{26}C_2 = \frac{26!}{2! 24!} \times \frac{26!}{2! 24!} = 325 \times 325 = 105625 \) ans.

6. 4 cards are of the same colour:
There are 2 cases: either all 4 cards are red, or all 4 are black.
4 red cards out of 26 red cards can be selected in \( = {}^{26}C_4 \) ways.
4 black cards out of 26 black cards can be selected in \( = {}^{26}C_4 \) ways.
\( \therefore \) Required number of ways of selection \( = {}^{26}C_4 + {}^{26}C_4 = \frac{26!}{4! 22!} + \frac{26!}{4! 22!} = 14950 + 14950 = 29900 \) ans.

Question. A group consisting of 4 girls & 7 boys. In how many ways can 5 members be selected such that the team consists of:
1. No girls.
2. At least 3 boys.
3. At most 2 boys.
4. At least 1 boy & 1 girl.
5. At most 1 girl is chosen.
6. A particular boy & a particular girl is always chosen.
Answer:
1. No girls:
Since no girls are to be selected, the 5 members must be selected from the 7 boys.
This can be selected in \( = {}^7C_5 \) ways \( = {}^7C_5 = {}^7C_2 = \frac{7 \times 6}{2} = 21 \) ans.

2. At least 3 boys:
There are three possible cases:
- Case I: Selecting 3 boys & 2 girls can be done in \( = {}^7C_3 \times {}^4C_2 \) ways \( = 35 \times 6 = 210 \).
- Case II: Selecting 4 boys & 1 girl can be done in \( = {}^7C_4 \times {}^4C_1 \) ways \( = 35 \times 4 = 140 \).
- Case III: Selecting 5 boys & 0 girls can be done in \( = {}^7C_5 \times {}^4C_0 \) ways \( = 21 \times 1 = 21 \).
\( \therefore \) Required number of ways of selection \( = 210 + 140 + 21 = 371 \) ans.

3. At most 2 boys:
Since there are only 4 girls available, we cannot choose 5 girls (0 boys). Thus, we have two possible cases:
- Case I: Selecting 2 boys & 3 girls can be selected in \( = {}^7C_2 \times {}^4C_3 \) ways \( = 21 \times 4 = 84 \).
- Case II: Selecting 1 boy & 4 girls can be selected in \( = {}^7C_1 \times {}^4C_4 \) ways \( = 7 \times 1 = 7 \).
\( \therefore \) Required number of ways of selection \( = 84 + 7 = 91 \) ans.

4. At least 1 boy & 1 girl:
The possible cases are:
- Case I: Selecting 1 boy & 4 girls can be selected in \( = {}^7C_1 \times {}^4C_4 \) ways \( = 7 \times 1 = 7 \).
- Case II: Selecting 2 boys & 3 girls can be selected in \( = {}^7C_2 \times {}^4C_3 \) ways \( = 21 \times 4 = 84 \).
- Case III: Selecting 3 boys & 2 girls can be selected in \( = {}^7C_3 \times {}^4C_2 \) ways \( = 35 \times 6 = 210 \).
- Case IV: Selecting 4 boys & 1 girl can be selected in \( = {}^7C_4 \times {}^4C_1 \) ways \( = 35 \times 4 = 140 \).
\( \therefore \) Required number of ways of selection \( = 7 + 84 + 210 + 140 = 441 \) ans.

5. At most 1 girl is chosen:
There are two possible cases:
- Case I: Selecting 4 boys & 1 girl can be selected in \( = {}^7C_4 \times {}^4C_1 \) ways \( = 35 \times 4 = 140 \).
- Case II: Selecting 5 boys & no girls can be selected in \( = {}^7C_5 \times {}^4C_0 \) ways \( = 21 \times 1 = 21 \).
\( \therefore \) Required number of ways of selection \( = 140 + 21 = 161 \) ans.

6. A particular boy & a particular girl is always chosen:
- Let the particular boy be A and the particular girl be B. They are selected in only 1 way.
- From the remaining 9 persons (3 girls and 6 boys), we now have to select the remaining 3 team members.
This can be selected in \( = {}^9C_3 \) ways \( = \frac{9 \times 8 \times 7}{3 \times 2 \times 1} = 84 \) ans.

Question. A polygon has \( n \) sides. Find the number of diagonals?
Answer:
(i) A polygon having \( n \) sides has \( n \) vertices.
(ii) Total number of lines that can be drawn using these \( n \) vertices \( = {}^nC_2 \).
(iii) These \( {}^nC_2 \) lines also contain the \( n \) sides of the polygon.
(iv) \( \therefore \) The number of diagonals \( = {}^nC_2 - n = \frac{n(n-1)}{2} - n = \frac{n^2 - n - 2n}{2} = \frac{n^2 - 3n}{2} \) ans.

Question. A polygon has 44 diagonals. Find the number of sides?
Answer:
(i) We know that the total number of diagonals having \( n \)-sides \( = \frac{n^2 - 3n}{2} \).
(ii) Given: number of diagonals \( = 44 \).
\( \therefore \frac{n^2 - 3n}{2} = 44 \)
\( \Rightarrow n^2 - 3n - 88 = 0 \)
\( \Rightarrow (n - 11)(n + 8) = 0 \)
\( \Rightarrow n = 11 \) or \( n = -8 \)
Since the number of sides cannot be negative, \( n = -8 \) is rejected.
\( \therefore \) There are 11 sides in the polygon.

Question. There are 10 points in a plane, out of which 4 points are collinear. Find no. of straight lines & no. of triangles?
Answer:
1. Total no. of straight lines:
(i) Total number of straight lines formed by joining 10 points \( = {}^{10}C_2 \).
(ii) However, 4 collinear points, when joined pairwise, give only 1 straight line instead of \( {}^4C_2 \) lines.
(iii) \( \therefore \) Required number of straight lines \( = {}^{10}C_2 - {}^4C_2 + 1 = 45 - 6 + 1 = 40 \) ans.

2. Total no. of triangles:
(i) Total number of triangles formed using 10 points \( = {}^{10}C_3 \).
(ii) However, the 4 collinear points cannot form any triangle.
(iii) \( \therefore \) Required number of triangles \( = {}^{10}C_3 - {}^4C_3 = 120 - 4 = 116 \) ans.

Question. There are ‘m’ no. of horizontal parallel lines & ‘n’ no. of vertical parallel lines. How many no. of parallelogram can be formed?
Answer:
(i) To form a parallelogram, we require two horizontal lines and two vertical lines.
(ii) Two horizontal lines out of 'm' horizontal parallel lines can be selected in \( = {}^mC_2 \) ways.
(iii) Two vertical lines out of 'n' vertical parallel lines can be selected in \( = {}^nC_2 \) ways.
(iv) \( \therefore \) Required number of parallelograms \( = {}^mC_2 \times {}^nC_2 \).

Question. From a class of 25 students, 10 are to be chosen for a party. There are 3 students who decide that either all of them will join or none of them will join. In how many ways can they be chosen?
Answer: There are two possible cases:

Case 1: Three particular students join the party:
(i) Now we have to select 7 students from the remaining 22 students.
(ii) This can be selected in \( {}^{22}C_7 \) ways.

Case 2: Three particular students do not join the party:
(i) Now we have to choose 10 students from the remaining 22 students.
(ii) This can be selected in \( {}^{22}C_{10} \) ways.

\( \therefore \) Required number of ways of selection \( = \text{Case 1} + \text{Case 2} = {}^{22}C_7 + {}^{22}C_{10} \)
\( = \frac{22!}{7! 15!} + \frac{22!}{10! 12!} = 170544 + 646646 = 817190 \) ans.

Question. A boy has 3 library tickets and 8 books of his interest in the library of these 8 books; he does not want to borrow chemistry part 2, unless chemistry part 1 is also borrowed. In how many ways can he choose the three books?
Answer: There are 2 cases:

Case 1: When chemistry part 1 is borrowed:
(i) Now, he has to select 2 more books out of the remaining 7 books.
(ii) This can be selected in \( = {}^7C_2 \) ways.

Case 2: When chemistry part 1 is not borrowed:
(i) Since chemistry part 1 is not borrowed, he cannot borrow chemistry part 2.
(ii) Now, he has to select 3 books out of the remaining 6 books.
(iii) This can be selected in \( {}^6C_3 \) ways.

\( \therefore \) Required number of ways of selection \( = \text{Case 1} + \text{Case 2} = {}^7C_2 + {}^6C_3 = 21 + 20 = 41 \) ans.

Question. A box contains 5 red balls & 5 black balls. In how many ways can 6 balls be selected such that:
1. There are exactly 2 red balls.
2. At least 3 red balls.
3. At least 2 red balls.
4. At least 2 balls from each colour.
5. No. of black balls & no. of red balls are equal.
6. Red balls are in majority.
Answer: We have 5 red balls and 5 black balls. We want to select 6 balls:

1. Exactly 2 red balls: We select 2 red balls and 4 black balls.
\( = {}^5C_2 \times {}^5C_4 = 10 \times 5 = 50 \) ways.

2. At least 3 red balls: We can select 3, 4, or 5 red balls.
- 3 Red, 3 Black: \( {}^5C_3 \times {}^5C_3 = 100 \)
- 4 Red, 2 Black: \( {}^5C_4 \times {}^5C_2 = 50 \)
- 5 Red, 1 Black: \( {}^5C_5 \times {}^5C_1 = 5 \)
\( \therefore \) Required ways \( = 100 + 50 + 5 = 155 \) ways.

3. At least 2 red balls: We can select 2, 3, 4, or 5 red balls.
Required ways \( = 50 \text{ (from part 1)} + 155 \text{ (from part 2)} = 205 \) ways.

4. At least 2 balls from each colour: Possible combinations are (2 Red, 4 Black), (3 Red, 3 Black), or (4 Red, 2 Black).
Required ways \( = 50 + 100 + 50 = 200 \) ways.

5. Number of black balls & number of red balls are equal: We must select 3 red and 3 black balls.
Required ways \( = {}^5C_3 \times {}^5C_3 = 10 \times 10 = 100 \) ways.

6. Red balls are in majority: We must select 4 or 5 red balls.
Required ways \( = 50 \text{ (4 Red, 2 Black)} + 5 \text{ (5 Red, 1 Black)} = 55 \) ways.

Question. If \( {}^{2n}C_3 : {}^nC_3 = 11:1 \), find n?
Answer: We have, \( \frac{{}^{2n}C_3}{{}^nC_3} = \frac{11}{1} \)
\( \Rightarrow \frac{\frac{(2n)!}{3!(2n-3)!}}{\frac{n!}{3!(n-3)!}} = 11 \)
\( \Rightarrow \frac{(2n)!(n-3)!}{(2n-3)! n!} = 11 \)
\( \Rightarrow \frac{2n(2n-1)(2n-2)(2n-3)!(n-3)!}{(2n-3)! \cdot n(n-1)(n-2)(n-3)!} = 11 \)
\( \Rightarrow \frac{2n(2n-1) \cdot 2(n-1)}{n(n-1)(n-2)} = 11 \)
\( \Rightarrow \frac{4(2n-1)}{n-2} = 11 \)
\( \Rightarrow 8n - 4 = 11n - 22 \)
\( \Rightarrow 3n = 18 \)
\( \Rightarrow n = 6 \) ans.

Question. If \( {}^{2n}C_3 : {}^nC_2 = 44:3 \), find n?
Answer: We have, \( \frac{{}^{2n}C_3}{{}^nC_2} = \frac{44}{3} \)
\( \Rightarrow \frac{\frac{2n(2n-1)(2n-2)}{3 \times 2 \times 1}}{\frac{n(n-1)}{2 \times 1}} = \frac{44}{3} \)
\( \Rightarrow \frac{2n(2n-1) \cdot 2(n-1)}{6} \times \frac{2}{n(n-1)} = \frac{44}{3} \)
\( \Rightarrow \frac{4(2n-1)}{3} = \frac{44}{3} \)
\( \Rightarrow 4(2n-1) = 44 \)
\( \Rightarrow 2n - 1 = 11 \)
\( \Rightarrow 2n = 12 \)
\( \Rightarrow n = 6 \) ans.