In a high school, a committee has to be formed from a group of 6 boys M1, M2, M3, M4, M5, M6 and 5 girls G1, G2, G3, G4, G5. (i) Let α1 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 α2 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 α3 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 α4 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 and such that both M1 and G1 are NOT in the committee together. LIST - I LIST - II P. The value of α1 is 1. 136 Q. The value of α2 is 2. 189 R. The value of α3 is 3. 192 S. The value of α4 is 4. 200 5. 381 6. 461 The correct option is P→1; Q→4; R→2; S→3 P→4; Q→6; R→2; S→1 P→4; Q→6; R→5; S→2 P→4; Q→2; R→3; S→1<p></p>

In a high school, a committee has to be formed from a group of 6 boys M1, M2, M3, M4, M5, M6 and 5 girls G1, G2, G3, G4, G5. (i) Let α1 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 α2 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 α3 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 α4 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 and such that both M1 and G1 are NOT in the committee together. LIST - I LIST - II P. The value of α1 is 1. 136 Q. The value of α2 is 2. 189 R. The value of α3 is 3. 192 S. The value of α4 is 4. 200 5. 381 6. 461 The correct option is P→1; Q→4; R→2; S→3 P→4; Q→6; R→2; S→1 P→4; Q→6; R→5; S→2 P→4; Q→2; R→3; S→1

(1) α1 = (6C3)(5C2) = 20 × 10 = 200 So P→4
(2) α2 = (6C1)(5C1) + (6C2)(5C2) + (6C3)(5C3) + (6C4)(5C4) + (6C5)(5C5) = 30 + 150 + 200 + 75 + 6 = 461 So Q→6
(3) α3 = (5C2)(6C3) + (5C3)(6C2) + (5C4)(6C1) + (5C5)(6C0) = 200 + 150 + 30 + 1 = 381 So R→5
(4) α4 = (5C2)(6C2) - (4C1)(5C1) + (5C3)(6C1) - (4C2)(5C0) + (5C4) = 150 - 20 + 100 - 6 + 5 = 229
(5C2)(4C2) = 60
(5C3)(4C1) = 40
(5C2)(4C2) + (5C3)(4C1) + 5C4 = 60 + 40 + 5 = 105
Total ways = 105
Total ways to select 4 members = 11C4 = 330
Total number of ways to select 4 members with at least 2 girls = 11C4- (6C4)(5C0) - (6C3)(5C1) - (6C2)(5C2) = 330 - 15 - 100 -150 = 65
(5C2)(4C2) = 60
(5C3)(4C1) = 40
α4 = (5C2)(4C2) + (5C3)(4C1) + 5C4 = 60 + 40 + 5 = 105
The number of ways to choose 4 members such that at least 2 are girls = (5C2)(6C2) + (5C3)(6C1) + (5C4)(6C0) = 150 + 100 + 5 = 255
Ways to choose M1 and G1 together = (4C1)(4C1) = 16
Then 255 - 16 = 239
α4 = 189
So S→2

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