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?

The correct answer is 21816.

We have 4 boxes, each with 3 red and 2 blue balls (all distinct). We must choose 10 balls total, with at least 1 red and 1 blue from each box.

Step 1: Determine possible ball distributions across boxes.

Let $n_i$ = number of balls chosen from box i. We need:

$n_1+n_2+n_3+n_4=10$, with each $n_i\ge 2$ (need at least 1 red + 1 blue) and each $n_i\le 5$.

Let $m_i=n_i-2$, so $m_1+m_2+m_3+m_4=2$, where $0\le m_i\le 3$.

The only distributions of 2 among 4 boxes are:

• Case 1: One box gets 2 extra → distribution is (4, 2, 2, 2) in some order.
• Case 2: Two boxes get 1 extra each → distribution is (3, 3, 2, 2) in some order.

Step 2: Count ways for a box with exactly 2 balls chosen.

Must pick exactly 1 red (from 3) and 1 blue (from 2):

$C(3,1)\times C(2,1)=3\times 2=6$

Step 3: Count ways for a box with exactly 3 balls chosen (at least 1 red, 1 blue).

Total ways to choose 3 from 5:

$C(5,3)=10$

Subtract invalid cases (0 blue, i.e., all 3 red from only 3 red balls):

$C(3,3)\times C(2,0)=1$

(0 red is impossible since we'd need 3 blue but only 2 exist.)

Valid ways: $10-1=9$

Step 4: Count ways for a box with exactly 4 balls chosen (at least 1 red, 1 blue).

Total ways to choose 4 from 5:

$C(5,4)=5$

Invalid cases: 0 blue would require 4 red, but only 3 exist. 0 red would require 4 blue, but only 2 exist. Both are impossible.

Valid ways: $5$

Step 5: Compute Case 1 — distribution (4, 2, 2, 2).

Choose which box gets 4 balls: $C(4,1)=4$ ways.

Ways = $4\times 5\times 6\times 6\times 6=4\times 5\times 216=4320$

Step 6: Compute Case 2 — distribution (3, 3, 2, 2).

Choose which 2 boxes get 3 balls: $C(4,2)=6$ ways.

Ways = $6\times 9\times 9\times 6\times 6=6\times 81\times 36=6\times 2916=17496$

Step 7: Total count.

$4320+17496=21816$

Therefore, the number of ways to choose 10 balls with at least one red and one blue from each box is 21816.