IPMAT 2024 - Quantitative Aptitude Question 32
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IPMAT 2024 - Quantitative Aptitude
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.