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Get access to the detailed solutions to the previous years questions asked in IIM IPMAT exam
Let us give one balloon to each one of them.
So, we are now left with 12 balloons to be distributed.
CASE I:
G1 G2 G3 G4 G5 B1 B1 B1
Giving no balloons to any boy and distributing remaining 12 balloons to the 5 girls.
Total ways = 12+5−1C5−1 = 16C4 = 1820 ways.
[By using the formula r−1Cn+r−1]
CASE II:
Giving 1 balloon to one of the boys, which can be done in 3 ways
And distributing remaining 11 balloons to the 5 girls, such that each girl should get at least one balloon.
This can be done in 10C4 = ways.
∴ Total ways 3 × 10C4 = 3 × 210 = 630 ways.
CASE III:
Giving 1 balloon each to 2 boys, which again can be done in 3 ways.
And distributing remaining 10 balloons to the 5 girls, such that each girl should get at least one balloon.
This can be done in 9C4 ways.
∴ Total ways 3 × 9C4 = 3 × 126 = 378 ways.
CASE IV:
Given 1 balloon to each of the 3 boys and distributing remaining 9 balloons to the 5 girls such that each girl should get at least one balloon.
∴ Total way = 1 × 8C4 = 70 way
CASE V:
Giving 2 balloon to exactly one boy which can be done in 3 ways and distributing remains 10 balloons to the 5 girls such that each get at least 2 balloons which can be done in only 1 way.
∴ Total ways = 3 × 1 = 3 ways
Adding all the possible ways obtained in all the 5 cases, we get
1820 + 630 + 378 + 70 + 3 = 2901 ways.