IPMAT Indore 2024 - QA (MCQ) Q32 Explanation

Explanation:

(*) None of these*

Note: In year 2024, the IIM Indore authorities accepted the objection raised by student’s community only for this particular question and granted 4 marks to all the test takers. They acknowledged that the question is alright but all the option given were incorrect. Though they did not share the correct answer for this question.

Don’t worry, SG has solved it for you.

Sol. 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 n+r−1 C 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 x 1 = 3 ways

Adding all the possible ways obtained in all the 5 cases, we get

1820 + 630 + 378 + 70 + 3 = 2901 ways. Ans.