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JEE Advanced 2018Paper 2
Exam
JEE Advanced
Year
2018
Paper
Paper 2
Subject
Mathematics
Chapter
Permutation and Combination

JEE Advanced 2018 · Paper 2 · Mathematics · Permutation and Combination

In a high school, a committee has to be formed from a group of 6...

In a high school, a committee has to be formed from a group of 66 boys
M1,M2,M3,M4,M5,M6{M}_{1}, {M}_{2}, {M}_{3}, {M}_{4}, {M}_{5}, {M}_{6} and 55 girls G1,G2,G3,G4,G5{G}_{1}, {G}_{2}, {G}_{3}, {G}_{4}, {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 55 members, having exactly 33 boys and 22 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 22 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 55 members, at least 22 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 44 members, having at least 22 girls and such that both M1{M}_{1} and G1{G}_{1} are NOT in the committee together.
LIST-I LIST-II
A. The value of α1{\alpha}_{1} is P. 136136
B. The value of α2{\alpha}_{2} is Q. 189189
C. The value of α3{\alpha}_{3} is R. 192192
D. The value of α4{\alpha}_{4} is S. 200200
T. 381381
U. 461461
The correct option is:

Options

  1. Aa-s;b-u;c-t;d-q;
  2. Ba-u;b-t;c-q;d-p;
  3. Ca-t;b-r;c-s;d-p;
  4. Da-t;b-r;c-q;d-p;

Correct answer

A. a-s;b-u;c-t;d-q;

Step-by-step solution

\left(1\right){\alpha}_{1} = \left(63\begin{array}{ll}6 \\ 3\end{array}\right)\left(52\begin{array}{ll}5 \\ 2\end{array}\right) = 200
So P→4P \to 4
\left(2\right){\alpha}_{2} = \left(61\begin{array}{ll}6 \\ 1\end{array}\right)\left(51\begin{array}{ll}5 \\ 1\end{array}\right) + \left(62\begin{array}{ll}6 \\ 2\end{array}\right)\left(52\begin{array}{ll}5 \\ 2\end{array}\right) + \left(63\begin{array}{ll}6 \\ 3\end{array}\right)\left(23\begin{array}{ll}2 \\ 3\end{array}\right) + \left(64\begin{array}{ll}6 \\ 4\end{array}\right)\left(54\begin{array}{ll}5 \\ 4\end{array}\right) + \left(65\begin{array}{ll}6 \\ 5\end{array}\right)\left(55\begin{array}{ll}5 \\ 5\end{array}\right)
= \left(115\begin{array}{ll}11 \\ 5\end{array}\right) - 1 = 461
So Q→6Q \to 6
\left(3\right){\alpha}_{3} = \left(52\begin{array}{ll}5 \\ 2\end{array}\right)\left(63\begin{array}{ll}6 \\ 3\end{array}\right) + \left(53\begin{array}{ll}5 \\ 3\end{array}\right)\left(62\begin{array}{ll}6 \\ 2\end{array}\right) + \left(54\begin{array}{ll}5 \\ 4\end{array}\right)\left(61\begin{array}{ll}6 \\ 1\end{array}\right) + \left(55\begin{array}{ll}5 \\ 5\end{array}\right)\left(60\begin{array}{ll}6 \\ 0\end{array}\right)
= \left(115\begin{array}{ll}11 \\ 5\end{array}\right) - \left(50\begin{array}{ll}5 \\ 0\end{array}\right)\left(65\begin{array}{ll}6 \\ 5\end{array}\right) - \left(51\begin{array}{ll}5 \\ 1\end{array}\right)\left(64\begin{array}{ll}6 \\ 4\end{array}\right) = 381
So R→5R \to 5
\left(4\right){\alpha}_{2} = \left(52\begin{array}{ll}5 \\ 2\end{array}\right)\left(62\begin{array}{ll}6 \\ 2\end{array}\right) - \left(41\begin{array}{ll}4 \\ 1\end{array}\right)\left(51\begin{array}{ll}5 \\ 1\end{array}\right) + \left(53\begin{array}{ll}5 \\ 3\end{array}\right)\left(61\begin{array}{ll}6 \\ 1\end{array}\right) - \left(42\begin{array}{ll}4 \\ 2\end{array}\right)\left(11\begin{array}{ll}1 \\ 1\end{array}\right) + \left(54\begin{array}{ll}5 \\ 4\end{array}\right) = 189
So S→2S \to 2
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