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 ?

Permutations and Combinations PYQ — JEE Advanced 2022 · Shift 2 · Q33 with Solution

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Correct answer: A

There are $4$ boxes. Each box has:

So each box contains $5$ distinct balls, and overall there are $20$ distinct balls.

We must choose exactly $10$ balls such that from each box, at least one red and one blue ball are chosen.

Let us see how many balls can be selected from a single box under the condition that at least one red and one blue must be chosen.

Since a box has only $2$ blue balls, and we need at least one blue and at least one red, the allowed selections from one box are:

Now we need a total of $10$ balls from $4$ boxes.

Let $x_i$ be the number chosen from box $i$ . Then $x_1+x_2+x_3+x_4=10,$ with each $x_i\ge 2$ .

Let $y_i=x_i-2\ge 0$ . Then $y_1+y_2+y_3+y_4=2.$

So the possible distributions of extra balls are:

Thus only two types of boxwise distributions are possible:

Choose which box contributes $4$ balls: $\binom{4}{1}=4.$

Now count selections:

From a box contributing $4$ balls with at least one red and one blue: $\binom{5}{4}-\binom{3}{4}-\binom{2}{4}=\binom{5}{4}=5,$ but better directly: $\binom{3}{3}\binom{2}{1}+\binom{3}{2}\binom{2}{2}=1\cdot 2+3\cdot 1=5.$
From a box contributing $2$ balls with at least one red and one blue, it must be exactly $1R+1B$ : $\binom{3}{1}\binom{2}{1}=6.$

So total for this case: $4\times 5\times 6^3 = 4\times 5\times 216=4320.$

Choose which $2$ boxes contribute $3$ balls: $\binom{4}{2}=6.$

Now count selections:

From a box contributing $3$ balls with at least one red and one blue: possible compositions are $2R+1B$ or $1R+2B$ . Hence $\binom{3}{2}\binom{2}{1}+\binom{3}{1}\binom{2}{2}=3\cdot 2+3\cdot 1=9.$
From a box contributing $2$ balls with at least one red and one blue: $\binom{3}{1}\binom{2}{1}=6.$

So total for this case: $\binom{4}{2}\times 9^2\times 6^2 = 6\times 81\times 36.$ Now, $81\times 36=2916,$ so $6\times 2916=17496.$

Adding both cases, $4320+17496=21816.$

Thus the required number of ways is $\boxed{21816}.$

So the correct option is A.