This page contains Combination Questions and Answers to help student get a clear understanding about variety of Questions based on Combinations and its familiar Concepts.

### Rules for Combination Questions and Answers

• We use combinations if an issue arise for the number of tricks of choosing things along with the series of choice is not to be considered.
• The values of the given combinations of n factors, reserved r at a time is:
$^nC_{r} = \frac{n!}{(r!)(n-r!)} = n (n-1)(n-2)… \frac{\text{ to r factors }}{r!}$
• While solving combinations, when n = r, the value of combinations is constantly equal to 1.

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### Sample Combination Questions and Answers

1. A man has seven blue marbles and three white marbles. He randomly takes out three marbles. Find the ways in which three marbles are of the same color are selected?

41

41

7.91%

36

36

77.97%

91

91

5.08%

120

120

9.04%

Total marbles = 7 + 3 =10

Total no. of ways for taking our 3 marbles out of 10 = 10C3

= 10!/3!*7!

=120

Event of taking out three marbles of similar color = 7C3 + 3C3

=[7!/(3!*4!)]+[3!/(3!*0!)]

=35+1

=36

2. A team has 12 players. The ways in which:

(a)  6 players can be selected to start the game?

(b)  3 starting guards?

924 and 220

924 and 220

83.87%

1011 ans 120

1011 ans 120

5.81%

950 and 350

950 and 350

7.1%

550 and 150

550 and 150

3.23%

Total no. of players = 12

Select 6 players out of 12  = 12C6

= 924

1. Out of 12, three players as starting guards = 12C3 = 220

3. In a group, there are 15 boys and 25 girls. Four players are chosen randomly. The ways in which 3 girls and 1 boy are chosen is:

31250

31250

6.25%

32200

32200

11.81%

34500

34500

67.36%

None of the above

None of the above

14.58%

Let us assume S as the sample space

And E is the event for choosing 3 girls and 1 boy

n(E) = (15C1 × 25C3)

= 15 * (25*24*23)/3*2*1

=15*2300

=34500

4. In a school team, there are 14 boys and 9 girls. A new team has to be formed for the inter-school tournament. To form a new team comprising of five students are to be selected such that there are at least 3 boys in the new team. Find the total ways in which it can be done?

24115

24115

68.1%

24151

24151

10.34%

21415

21415

11.21%

25411

25411

10.34%

The new team may comprise both (3 boys and 2 girls), or it may have (4 boys and 1 girl) or (only 5 boys).

Total required ways = (14C3 x 9C2) + (14C4 x 9C1) + (14C5)

= (13104 + 9009 + 2002)

= 24115 ways.

5. Find the ways in which a group, containing 4 men and 5 women, which can be made from 9 men and 10 women?

12367

12367

8.26%

38152

38152

13.22%

31752

31752

73.55%

37500

37500

4.96%

Required ways

= (9C5 x 10C5)

= 31752

6. Find the ways in which a group, containing 4 men and 5 women, which can be made from 9 men and 10 women?

12367

12367

3.48%

38152

38152

11.3%

31752

31752

80.87%

37500

37500

4.35%

Required ways

= (9C5 x 10C5)

= 31752

7. In a bag, there are 2 pink beads, 3 black beads, and 4 red beads. Calculate the number of ways one can draw 3 beads from the bag, such that at least one out of the three beads is a black bead?

24

24

11.82%

32

32

9.09%

64

64

74.55%

144

144

4.55%

Beads picked can be = (1 black and 2 pink/red), (2 black and 1 pink/red), or (3 black).

Total ways the beads can be chosen = (3C1 x 6C2) + (3C2 x 6C1) + (3C3)

= (45 + 18 + 1)

= 64 ways

8. In a club, there are 12 men and 9 women. To form a new team of eight people the leader wants at least 4 men in the new team. Find the ways in which it can be done?

16300

16300

4.95%

12300

12300

10.89%

163926

163926

74.26%

15125

15125

9.9%

The new team may comprise either (4 men and 4 women) or it may have (5 men and 3 women) or (6 men and 2 women) or (7 men and 1 woman) or ( all 8 men).

Total ways = (12C4 x 9C4) + (12C5 x 9C3) + (12C6 x 9C2) + (12C7 x 9C1) (12C8)

= (495 * 126) + (729 * 84) + (924 * 36) + (792 * 9) + (495)

= (62370 + 61236 + 33264 + 6561 + 495)

= 163926 ways.

9. In a basket, there are 2 oranges, 3 apples, and 4 bananas. In how many ways can a person draw 3 fruits from the basket, such that at least two out of the three fruits are bananas?

34

34

72.82%

72

72

13.59%

54

54

9.71%

60

60

3.88%

Fruits picked can be = (2 bananas and 1 not a banana) or (all 3 bananas).

Total ways fruits can be chosen = (4C2 x 5C1) + (4C3)

= 6 * 5 + 4

= 30 + 4

= 34 ways

10. Amongst 6 boys and 4 girls, a new team has to be formed comprising of four people. In how many ways the team can be formed such that at least one girl is there?

195

195

76.36%

200

200

8.18%

105

105

9.09%

150

150

6.36%

Amongst 6 boys and 4 girls, four students are to be carefully chosen such that out of 4, there is at least one girl.

There are cases 4 cases possible

Case 1: Select 4 girls
Total ways for it = 4C4

Case 2: Select 3 girls and 1 boy
Total ways for it = 4C3 × 6C1

Case 3: Select 2 boys and 2 girls
Total ways for it = 6C2 × 4C2

Case 4: Select 3 boys and 1 girl
Total ways for it = 6C3 × 4C1

Total ways
4C4 + 4C3 × 6C16C2 × 4C2 + 6C3 × 4C1
= 1+ 24+ 90 + 80

= 195 ways

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