Problems on Ages
Problems on Ages - 1 Problems on Ages - 2 Problems on Ages - 3 Problems on Ages - 4Problems on Average
Problems on Average - 1 Problems on Average - 2 Problems on Average - 3Odd Man Out Series
Odd Man Out Series - 1 Odd Man Out Series - 2 Odd Man Out Series - 3 Odd Man Out Series - 4Problems on Percentage
Problems on Percentage - 1 Problems on Percentage - 2 Problems on Percentage - 3 Problems on Percentage - 4 Problems on Percentage - 5Permutation and Combination
Permutation and Combination - 1 Permutation and Combination - 2 Permutation and Combination - 3Problems on Probability
Problems on Probability - 1 Problems on Probability - 2 Problems on Probability - 3 Problems on Probability - 4Problems on Profit Loss
Profit and Loss - 1 Profit and Loss - 2 Profit and Loss - 3 Profit and Loss - 4 Profit and Loss - 5 Profit and Loss - 6Problems on Test Divisibility
Test Divisibility - 1 Test Divisibility - 2 Test Divisibility - 3Time and Work
Time and Work - 1 Time and Work - 2 Time and Work - 3The ones who are crazy enough to think they can change the world are the ones who do.- Steve Jobs
Tricky Questions on Permutation and Combination
Permutation and Combination - Aptitude
6.In how many ways, a committee of 5 members can be selected from 6 men and 5 ladies, consisting of 3 men and 2 ladies?
x
Option: D
Explanation
3 men out of 6 and 2 ladies out of 5 are to be chosen
Required number of words = 6C3*5C2 = ((6 * 5 * 4)/(3 * 2 * 1)) * ((5 * 4)/(2 * 1)) = 200
7.In how many ways can a group of 5 men and 2 women be made out of a total of 7 men and 3 women?
x
Option: A
Explanation
Required number of ways = 7C5*3C2=7C2*3C1 = ((7 * 6)/(2 * 1)) * 3 = 63
8.In how many ways a committee, consisting of 5 men and 2 women can be formed from 8 men and 10 women?
x
Option: B
Explanation
required number of ways = 8C5*10C6=8C3*10C4
=((8 * 7 * 6)/(3 * 2 * 1)) * ((10 * 9 * 8 * 7)/(4 * 3 * 2 * 1)) = 11760
9.From a group of 7 men and 6 women, five persons are to be selected to form a committee so that at least 3 men are there on the committee. In how many ways can it be done?
x
Option: C
Explanation
We may have 3 men and 2 women or 4 men and 1 women or 5 men only
Required number of words = (7C3 * 6C2) + (7C4 * 6C1) + 7C5
= (((7 * 6 * 5)/(3 * 2 * 1)) * ((6 * 5)/(2 * 1))) + (((7 * 6 * 5)/(3 * 2 * 1)) * 6) + ((7 * 6)/(2 * 1))
= 525 + 210 + 21 = 756
10.In a group of 6 boys and 4 girls, four children are to be selected. In how many different ways can they be selected such that at least one boy should be there?
x
Option: A
Explanation
We may have 1 boy and 3 girls or 2 boys and 2 girls or 3 boys and 1 girl or 4 boys
required number of ways = (6C1 * 4C3) + (6C2 * 4C2)+(6C3 * 4C1) + 6C4
= (6C1 * 4C1) + (6C2 * 4C2) + (6C3 * 4C1) + 6C2
= (6 * 4) + (((6 * 5)/(2 * 1)) * ((4 * 3)/(2 * 1))) + (((6 * 5 * 4)/(3 * 2 * 1)) * 4) + ((6 * 5)/(2 * 1))
= 24 + 90 + 80 + 15 = 209
Related to Problems on Permutation and Combination
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