- #1

hotvette

Homework Helper

- 988

- 3

## Homework Statement

How many one-pair 5-card poker hands are there in a standard 52-card deck?

## Homework Equations

[tex]

{}^nC_k = \frac{n!}{(n-k)!}

[/tex]

3. The Attempt at a Solution

I've seen the following solution in several different places:

[tex]

n = {}^{13}C_1 \cdot {}^{4}C_2 \cdot {}^{12}C_3 \cdot {}^{4}C_1 \cdot {}^{4}C_1 \cdot {}^{4}C_1 = 1098240

[/tex] and every term makes perfect sense. What I don't understand though, is why the following:

[tex] n = {}^{13}C4 \cdot {}^{4}C_2 \cdot {}^{4}C_1 \cdot {}^{4}C_1 \cdot {}^4C_1

[/tex] underestimates the answer by a factor of 4. To me 13 choose 4 makes just as much sense as 13 choose one followed by 12 choose three, but of course one is correct and the other isn't. Can someone explain?

Last edited: