JEE Advance - Mathematics (2018 - Paper 2 Offline - No. 16)
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 $$\alpha $$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 $$\alpha $$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.
i) Let $$\alpha $$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 $$\alpha $$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 M1 and G1 are NOT in the committee together.
The correct option is
(i) Let $$\alpha $$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 $$\alpha $$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.
i) Let $$\alpha $$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 $$\alpha $$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 M1 and G1 are NOT in the committee together.
| LIST-I | LIST-II |
|---|---|
| P. The value of $\alpha_1$ is | 1. 136 |
| Q. The value of $\alpha_2$ is | 2. 189 |
| R. The value of $\alpha_3$ is | 3. 192 |
| S. The value of $\alpha_4$ is | 4. 200 |
| 5. 381 | |
| 6. 461 |
P $$ \to $$ 4; Q $$ \to $$ 6; R $$ \to $$ 2; S $$ \to $$ 1
P $$ \to $$ 1; Q $$ \to $$ 4; R $$ \to $$ 2; S $$ \to $$ 3
P $$ \to $$ 4; Q $$ \to $$ 6; R $$ \to $$ 5; S $$ \to $$ 2
P $$ \to $$ 4; Q $$ \to $$ 2; R $$ \to $$ 3; S $$ \to $$ 1
Explanation
Given 6 boys M1, M2, M3, M4, M5, M6 and 5 girls G1, G2, G3, G4, G5
(i) $$\alpha $$1 $$ \to $$ Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls.
i.e., $${}^6{C_3} \times {}^5{C_2} = 20 \times 10 = 200$$
$$ \therefore $$ $$\alpha $$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
$$\alpha $$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 M1 and G1 are not included together.
G1 is included $$ \to $$
$${}^4{C_1}.{}^5{C_2} + {}^4{C_2}.{}^5{C_1} + {}^4{C_3}$$
40 + 30 + 4 = 74
M1 is included $$ \to $$
$${}^4{C_2}.{}^5{C_1} + {}^4{C_3}$$ = 30 + 4 = 34
G1 and M1 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.
(i) $$\alpha $$1 $$ \to $$ Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls.
i.e., $${}^6{C_3} \times {}^5{C_2} = 20 \times 10 = 200$$
$$ \therefore $$ $$\alpha $$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
$$\alpha $$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 M1 and G1 are not included together.
G1 is included $$ \to $$
$${}^4{C_1}.{}^5{C_2} + {}^4{C_2}.{}^5{C_1} + {}^4{C_3}$$
40 + 30 + 4 = 74
M1 is included $$ \to $$
$${}^4{C_2}.{}^5{C_1} + {}^4{C_3}$$ = 30 + 4 = 34
G1 and M1 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.
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