Question

Assertion :The number of ways in which 4 red, 3 yellow & 2 green discs be arranged in a row (if the discs of same colour are indistinguishable) is $1260.$ Reason: The number of permutation of n objects out of which $p_1$ are of first kind, $p_2$ are of second kind, .... $p_k$ are of $k^{th}$ kind & rest are different kind is $\frac{n!}{p_1!p_2!...P_k!}$

• A. Both (A) & (R) are True & (R) is correct explanation of (A),
• B. Both (A) & (R) are True but (R) is not the correct explanation of (A).
• C. (A)is True but (R) is False.
• D. (A)is False but (R) is True.

Answer 1

The correct option is A Both (A) & (R) are True & (R) is correct explanation of (A),

Total number of discs are $4+3+2=9,$ out of these 9 discs, 4 are of first kind (Red), 3 are of second kind (yellow), 2 are of third kind (green)

$\therefore$ Required number of arrangements $=\frac{9!}{4!3!2!}=1260.$

Statement B is correct explaination of statement A