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Mathematics · Permutations and Combinations

JEE Advanced 2018 — Paper 2 — Question 43

In a high school, a committee has to be formed from a group of 6 boys M1,M2,M3,M4,M5,M6M_{1}, M_{2}, M_{3}, M_{4}, M_{5}, M_{6} and 5

girls G1,G2,G3,G4,G5\mathrm{G}_{1}, \mathrm{G}_{2}, \mathrm{G}_{3}, \mathrm{G}_{4}, \mathrm{G}_{5}.

(i) Let α1\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 α2\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.

(iii) Let α3\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 α4\alpha_{4} be the total number of ways in which the committee can be formed such that the committee has 4 members, having atleast 2 girls and such that both M1M_{1} and G1G_{1} are NOT in the committee together.

LIST-ILIST-II
P. The value of α1\alpha_{1} is1. 136
Q. The value of α2\alpha_{2} is2. 189
R. The value of α3\alpha_{3} is3. 192
S. The value of α4\alpha_{4} is4. 200
5. 381
6. 461
  1. Option A:

    P→4;Q→6;R→2;S→1\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow \mathbf{6} ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow \mathbf{1}

  2. Option B:

    P→1;Q→4;R→2;S→3\mathrm{P} \rightarrow 1 ; \mathrm{Q} \rightarrow 4 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 3

  3. Option C:

    P→4;Q→6;R→5;S→2\mathrm{P} \rightarrow 4 ; \mathrm{Q} \rightarrow \mathbf{6} ; \mathrm{R} \rightarrow 5 ; \mathrm{S} \rightarrow 2

    Correct
  4. Option D:

    P→4;Q→2;R→3;S→1\mathbf{P} \rightarrow 4 ; \mathbf{Q} \rightarrow 2 ; R \rightarrow 3 ; S \rightarrow \mathbf{1}

Answer: C

Step-by-step solution

α1=6C3⋅5C2=20⋅10=200\alpha_{1}={ }^{6} \mathrm{C}_{3} \cdot{ }^{5} \mathrm{C}_{2}=20 \cdot 10=200 α1=6C1⋅5C1+6C25C2+6C35C3+6C45C4+6C55C5\alpha_{1}={ }^{6} \mathrm{C}_{1} \cdot{ }^{5} \mathrm{C}_{1}+{ }^{6} \mathrm{C}_{2}{ }^{5} \mathrm{C}_{2}+{ }^{6} \mathrm{C}_{3}{ }^{5} \mathrm{C}_{3}+{ }^{6} \mathrm{C}_{4}{ }^{5} \mathrm{C}_{4}+{ }^{6} \mathrm{C}_{5}{ }^{5} \mathrm{C}_{5} =30+150+200+75+6=30+150+200+75+6 =461=461 α3=5C26C3+5C36C2+5C46C1+5C5\alpha_{3}={ }^{5} \mathrm{C}_{2}{ }^{6} \mathrm{C}_{3}+{ }^{5} \mathrm{C}_{3}{ }^{6} \mathrm{C}_{2}+{ }^{5} \mathrm{C}_{4}{ }^{6} \mathrm{C}_{1}+{ }^{5} \mathrm{C}_{5} =200+150+30+1=381=200+150+30+1=381 α4=1⋅[4C1⋅5C2+4C25C1+4C3]⏟with G1+1⋅[4C25C1+4C3]⏟with M1+[4C4+4C3⋅5C1+4C2+5C2]⏟neither M1 nor G1\alpha_{4}=\underbrace{1 \cdot\left[{ }^{4} \mathrm{C}_{1} \cdot{ }^{5} \mathrm{C}_{2}+{ }^{4} \mathrm{C}_{2}{ }^{5} \mathrm{C}_{1}+{ }^{4} \mathrm{C}_{3}\right]}_{\text {with } \mathrm{G}_{1}}+\underbrace{1 \cdot\left[{ }^{4} \mathrm{C}_{2}{ }^{5} \mathrm{C}_{1}+{ }^{4} \mathrm{C}_{3}\right]}_{\text {with } \mathrm{M}_{1}}+\underbrace{\left[{ }^{4} \mathrm{C}_{4}+{ }^{4} \mathrm{C}_{3} \cdot{ }^{5} \mathrm{C}_{1}+{ }^{4} \mathrm{C}_{2}+{ }^{5} \mathrm{C}_{2}\right]}_{\text {neither } \mathrm{M}_{1} \text { nor } \mathrm{G}_{1}} =(40+30+4)+(34)+(1+20+60)=(40+30+4)+(34)+(1+20+60) =74+34+81=189=74+34+81=189 P→4,Q→6,R→5, S→2\mathrm{P} \rightarrow 4, \mathrm{Q} \rightarrow 6, \mathrm{R} \rightarrow 5, \mathrm{~S} \rightarrow 2.

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2018
Paper
Paper 2
Subject
Mathematics
Chapter
Permutations and Combinations
Topic
Combinations