Verified Solution • 2025–26 Session moderate

QJEEAdv18Maf2b8: JEE Advanced 2018 Mathematics - Class 11 JEEAdv18Maf2b8 Solution: In a high school, a committee has to be formed from a group of…

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 α\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 α\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 α\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 α\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 α1\alpha_1 is 1. 136
Q. The value of α2\alpha_2 is 2. 189
R. The value of α3\alpha_3 is 3. 192
S. The value of α4\alpha_4 is 4. 200
5. 381
6. 461
The correct option is

Multiple Choice 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

Step-by-Step Verified Solution

Given 6 boys M1, M2, M3, M4, M5, M6 and 5 girls G1, G2, G3, G4, G5

(i) α\alpha 1 →\to Total number of ways of selecting 3 boys and 2 girls from 6 boys and 5 girls.

i.e., 6C3×5C2=20×10=200{}^6{C_3} \times {}^5{C_2} = 20 \times 10 = 200

∴\therefore α\alpha 1 = 200

(ii) α\alpha 2 →\to Total number of ways selecting at least 2 member and having equal number of boys and girls i.e., 6C15C1+6C25C2+6C35C3+6C45C4+6C55C5{}^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

α\alpha 2 = 461

(iii) α\alpha 3 →\to Total number of ways of selecting 5 members in which at least 2 of them girls

i.e., 5C26C3+5C36C2+5C46C1+5C56C0{}^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= 200 + 150 + 30 + 1 = 381

α\alpha 3 = 381

(iv) α\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

4C1.5C2+4C2.5C1+4C3{}^4{C_1}.{}^5{C_2} + {}^4{C_2}.{}^5{C_1} + {}^4{C_3}

40 + 30 + 4 = 74

M1 is included →\to

4C2.5C1+4C3{}^4{C_2}.{}^5{C_1} + {}^4{C_3} = 30 + 4 = 34

G1 and M1 both are not included

4C4+4C3.5C1+4C2.5C2{}^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

α\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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