Wipro Company Numerical Ability Probability

Mr verma has 4 diff paintings that he wishes to divide among his 3 children. In how many ways can he do this if each child must get at least 1 painting and all paintings are diff?
Options--> 48,18,36,72

Read Solution (Total 4)

first child could get 1 out of 4 paintings: 4
second can get 1 out of remaining 3: 3
third will get 1 out of remaining 2: 2

now the last painting could be given to any of the three child in 3 diff ways:

thus total number of ways: (4 * 3 * 2) * 3
9 years agoHelpfull: Yes(12) No(0)

3(4C1*3C1*2C1)=72
9 years agoHelpfull: Yes(3) No(1)
36
4c2*2c1*1c1=12
therfore 12+12+12=36 ways
9 years agoHelpfull: Yes(0) No(3)
3childrens are arranged in 3!ways
4 different paintings are given to them in 4!ways
So,4!*3!=72
6 years agoHelpfull: Yes(0) No(2)

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