Possible draws in which exactly one player gets exactly one 8

Question

The card deck has 30 cards in three different colors. Cards of each color are numbered 1 to 10. Each player draws 4 cards. The order in which cards are drawn to players' hands is not important.

The question is in how many possible draws exist in which exactly one player receives exactly one card which has the number 8 on it. I thought that there are 2 players, for each player 3 options to draw the 8, $27 \choose 3$ for the three other cards for that player and another $24 \choose 4$ cards for the other player. This totals to 186,486,300 which is not the correct answer.

Any guidance would be appreciated.

Your answer looks right to me; what is the answer supposed to be? — saulspatz, Oct 07 '19 at 19:39
There are two possible outcomes (from what I am reading). 1: player 1 has an 8, player 2 has no 8. 2: player 1 has an 8, player 2 has 2x 8s. — SlipEternal, Oct 07 '19 at 20:04

score 4 · Accepted Answer · answered Oct 07 '19 at 20:16

Choose one player to have exactly one 8: $$\dbinom{2}{1}$$

Choose the 8: $$\dbinom{3}{1}$$

Choose three other non-8 cards for the player holding the 8 and four other non-8 cards for the other player:

$$\dbinom{27}{3}\dbinom{24}{4}$$

This gives the answer you show.

Now, assume one player has one 8 while the other player has two.

Choose the player with exactly one 8: $$\dbinom{2}{1}$$

Choose the singleton 8: $$\dbinom{3}{1}$$

The remaining two eights go to the other player. Choose 3 additional non-8 cards for the player with one 8 and two additional non-8 cards for the player with two 8's.

$$\dbinom{27}{3}\dbinom{24}{2}$$

Total (with addition principle):

$$\dbinom{2}{1}\dbinom{3}{1}\dbinom{27}{3}\left(\dbinom{24}{4}+\dbinom{24}{2}\right) = 191,330,100$$

Yup that's the one, forgot about the option where the other one has the two other 8's. Thanks! — HippoTerrific, Oct 08 '19 at 15:46

Possible draws in which exactly one player gets exactly one 8

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