Problems on Permutations and Combinations:
1. From a group of 4 men and 2 women, 3 persons are to be selected to form a committee so that at least 2 men are there on the committee. In how many ways can it be done?
Solution:
= 4C2 * 2C1 + 4C3
= 4C2 * 2C1 + 4C1
= (4*3)/(1*2) * 2 +
= 12 +4
= 16.
2. From a group of 6 boys and 3 girls, 5 children are to be selected. In how many different ways can they be selected such that at least 3 boys should be there?
Solution:
= ( 6C3 * 3C2) + ( 6C4 * 3C1 ) +( 6C5 )
= [(6*5*4)/(1*2*3) * 3C1] +[ 6C2 * 3 ] + 6C1
= [20*3] +[(6*5/1*2)*3] +6
= 60+45+6
= 111.
1. From a group of 4 men and 2 women, 3 persons are to be selected to form a committee so that at least 2 men are there on the committee. In how many ways can it be done?
Solution:
= 4C2 * 2C1 + 4C3
= 4C2 * 2C1 + 4C1
= (4*3)/(1*2) * 2 +
= 12 +4
= 16.
2. From a group of 6 boys and 3 girls, 5 children are to be selected. In how many different ways can they be selected such that at least 3 boys should be there?
Solution:
= ( 6C3 * 3C2) + ( 6C4 * 3C1 ) +( 6C5 )
= [(6*5*4)/(1*2*3) * 3C1] +[ 6C2 * 3 ] + 6C1
= [20*3] +[(6*5/1*2)*3] +6
= 60+45+6
= 111.
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