Consider 4 boxes,where each box contains 3 re | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) Let $n_i$ be the number of balls chosen from box $i$,where $n_i \ge 2$ and $\sum_{i=1}^4 n_i = 10$. Since each box must contain at least one red a

Vedclass · Accepted Answer

$21816$

Vedclass · Answer

(A) Let $n_i$ be the number of balls chosen from box $i$,where $n_i \ge 2$ and $\sum_{i=1}^4 n_i = 10$. Since each box must contain at least one red and one blue ball,the possible distributions of $(n_1, n_2, n_3, n_4)$ are permutations of $(4, 2, 2, 2)$ and $(3, 3, 2, 2)$. Case $I$: Distribution $(4, 2, 2, 2)$. Number of ways to choose $4$ balls from one box such that at least one red and one blue are chosen: $\binom{5}{4} - \binom{3}{4} - \binom{2}{4} = 5 - 0 - 0 = 5$. Number of ways to choose $2$ balls from one box such that at least one red and one blue are chosen: $\binom{3}{1} 	imes \binom{2}{1} = 3 	imes 2 = 6$. Total ways for this case: $\binom{4}{1} 	imes 5 	imes 6^3 = 4 	imes 5 	imes 216 = 4320$. Case $II$: Distribution $(3, 3, 2, 2)$. Number of ways to choose $3$ balls from one box such that at least one red and one blue are chosen: $\binom{5}{3} - \binom{3}{3} - \binom{2}{3} = 10 - 1 - 0 = 9$. Total ways for this case: $\binom{4}{2} 	imes 9^2 	imes 6^2 = 6 	imes 81 	imes 36 = 17496$. Total ways = $4320 + 17496 = 21816$.

Consider $4$ boxes,where each box contains $3$ red balls and $2$ blue balls. Assume that all $20$ balls are distinct. In how many different ways can $10$ balls be chosen from these $4$ boxes so that from each box at least one red ball and one blue ball are chosen?

Similar Questions

In an examination,a student has to write exams in $8$ different subjects. He is declared fail if he fails in at least one subject. The number of ways in which he can fail is

If ${}^n C_{r-1}=36$, ${}^n C_r=84$, and ${}^n C_{r+1}=126$, then the value of ${}^n C_8$ is

There are $5$ volumes of Mathematics among $25$ books. They are arranged on a shelf in random order. The probability that the volumes of Mathematics stand in increasing order from left to right (the volumes are not necessarily kept side by side) is

$A$ man has $7$ relatives: $4$ ladies and $3$ gentlemen. His wife also has $7$ relatives: $3$ ladies and $4$ gentlemen. In how many ways can they invite a dinner party of $3$ ladies and $3$ gentlemen,such that there are $3$ of the man's relatives and $3$ of the wife's relatives?

Let $S$ denote the set of $4$-digit numbers $abcd$ such that $a > b > c > d$ and $P$ denote the set of $5$-digit numbers having the product of its digits equal to $20$. Then $n(S) + n(P)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module