# Math Help - Answer in the back of the book wrong?

1. ## Answer in the back of the book wrong?

Can someone please verify my answer? Because I'm sure I'm doing this question right but the answer at the back differs.

Q: A bridge hand consists of 13 cards. How many bridge hands include 5 of one suit, 6 cards of a second, and 2 cards of a third?

Isn't the solution: 13C5 x 13C6 x 13c2 = 172262376?

The answer at the back is 4134297024

EDIT:

Oh I just realized that I have to account for the fact that only 3 of the 4 suits are being utilized. And order is important so do I multiply the 172 262 376 by 4P3?

2. Hello, MATHDUDE2!

Because I'm sure I'm doing this question right but the answer at the back differs.

Q: A bridge hand consists of 13 cards.
How many bridge hands include 5 of one suit, 6 cards of a second, and 2 cards of a third?

Isn't the solution: $_{13}C_5 \times _{13}C_6 \times _{13}C_2 \:=\: 172,262,376$ ?

The answer at the back is: $4,134,297,024$

EDIT:
Oh, I just realized that I have to account for the fact that only 3 of the 4 suits are being utilized.
And order is important, so do I multiply the 17, 262,376 by $_4P_3$? . . . . Yes!
Here's the way I approached it . . .

There are ${\color{blue}4}$ choices for the first suit.
. . There are: ${\color{blue}_{13}C_5}$ ways to get 5 cards of that suit.

There are ${\color{blue}3}$ choices for the second suit.
. . There are: ${\color{blue}_{13}C_6}$ ways to get 6 cards of that suit.

There are ${\color{blue}2}$ choices for the third suit.
. . There are: ${\color{blue}_{13}C_2}$ ways to get 2 cards of that suit.

Therefore, the number of ways is: . $\left(4 \times _{13}C_5\right) \times \left(3 \times _{13}C_6\right) \times \left(2 \times _{13}C_2\right)$