In a high school, a committee has to be formed from a group of 6 boys `M_(1), M_(2), M_(3), M_(4), M_(5), M_(6)` and 5 girls `G_(1), G_(2), G_(3), G_(4), G_(5)`.
(i) Let `a_(1)` be the total number of ways in which the committee can be formed such that the committee has 5 members, having exactly 3 boys and 2 girls
(ii) Let `a_(2)` be the total number of ways in which the committee can be formed such that the committee has at least 2 members, and having an equal number of boys and girls.
(iii) Let `a_(3)` be the total number of ways in which the committee can be formed such that the committee has 5 members, at least 2 of them being girls.
(iv) Let `a_(4)` be the total number of ways in which the committee can be formed such that the committee has 4 members, having at least 2 girls such that both `M_(1)` and `G_(1)` are NOT in the committee together.

The correct option is

Options

A. `P to 4, Q to 6, R to 2, S to 1`
B. `P to 1, Q to 4, R to 2, S to 3`
C. `P to 4, Q to 6, R to 5, S to 2`
D. `P to 4, Q to 2, R to 3, S to 1`

Answer

C

Explanation

Given 6 boys `M_(1), M_(2), M_(3), M_(4), M_(5), M_(6)` and 5 girls `G_(1), G_(2), G_(3), G_(4), G_(5)`
(i) `alpha_(1) to` Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls.
i.e., `""^(6)C_(3) xx ""^(5)C_(2) = 20 xx 10 = 200`
`therefore` `alpha_(1) = 200`
(ii) `alpha_(2) to` Total number of ways selecting at least 2 member and having equal number of boys and girls
i.e., `""^(6)C_(1) ""^(5)C_(1) + ""^(6)C_(2) ""^(5)C_(2) + ""^(6)C_(3) ""^(5)C_(3) + ""^(6)C_(4) ""^(5)C_(4) + ""^(6)C_(5) ""^(5)C_(5)`
= 30 + 150 + 200 + 75 + 6 = 461
`rArr" " alpha_(2) = 461`
(iii) `alpha_(3) to` Total number of ways of selecting 5 members in which at least 2 of them girls
i.e., `""^(5)C_(2) ""^(6)C_(3) + ""^(5)C_(3) ""^(6)C_(2) + ""^(5)C_(4) ""^(6)C_(1) + ""^(5)C_(5) ""^(6)C_(0)`
= 200 + 150 + 30 + 1 = 381
`alpha_(3) = 381`
(iv) `alpha_(4) to` Total number of ways of selecting 4 members in which at least two girls such that `M_(1)` and `G_(1)` are not included together.
`G_(1)` is included `to ""^(4)C_(1) * ""^(5)C_(2) + ""^(4)C_(2) * ""^(5)C_(1) + ""^(4)C_(3)`
= 40 + 30 + 4 = 74
`M_(1)` is included `to ""^(4)C_(2) * ""^(5*)C_(1) + ""^(4)C_(3) = 30 + 4 = 34`
`G_(1)` and `M_(1)` both are not included
`""^(4)C_(4) + ""^(4)C_(3) * ""^(5)C_(1) + ""^(4)C_(2) * ""^(5)C_(2)`
1 + 20 + 60 = 81
`therefore` Total number = 74 + 34 + 81 = 189
`alpha_(4) = 189`
Now `P to 4, Q to 6, R to 5, S to 2`
Hence, option (c) is correct.