# Why do I keep choosing the wrong probability rule?

I was wondering if someone could help clarify probability rules, although I think I understand them, whenever I have a straight forward and/or probability question, I seem to want to do a permutation or combination. For example...using the numbers 1-7 form a 4 digit number without repeats. When I saw this I immediately wanted to do a permutation of 7P4, only to find out that it is 7*6*5*4. I hope this isn't too simple or vague a question, thanks in advance.

• Well, ${}_7\text{P}_4$ is $(7)(6)(5)(4)$ so no harm done. I prefer $(7)(6)(5)(4)$, more concrete feeling. – André Nicolas May 9 '16 at 22:00
• Oh wow, I did nCr on my calculator thinking I was doing nPr. Since I got the wrong answer I just thought I didn't understand this at all hehe. Thanks guys, I have a stats final tomorrow; I think I'm a little bit more confident now. Emphasis on the I think. – mac_33 May 9 '16 at 22:06
• Good luck for tomorrow! – Peter May 9 '16 at 22:15

The value $_7P_4$ and $7 \cdot 6 \cdot 5 \cdot 4$ are the same.

The $P$ stands for permutations, which is exactly what you want, because $7612$ is distinct from $1267$.

If it were

$$_7C_4 = \frac{7!}{4!(7-4)!}$$

then these would be combinations, and would be incorrect for this question because the order of the digits does matter here.

But what you have is correct:

$$_7P_4 = \frac{7!}{(7-4)!} = 7 \cdot 6 \cdot 5 \cdot 4.$$