Question

There are 12 face cards in a standard deck of 52 cards. How many ways can you arrange a standard deck of 52 cards such that the first card is a face card?

A. 12P1
B. 12C1
C. 12P1 × 51P51
D. 12C1 × 51P51
E. 12C1 × 51C51

Answer 1

Hints:
(1)
There are 12 face cards in a deck.
How many ways can you arrange a single face card? There are 12 ways.
Since 12C1 is the same as 12P1 (equal to 12!/(12!0!) =12!/(12!) = 1), how do we know which one to use?
We can think of the hypothetical case of arranging two facecards from 12:
12 choices for the first, and 11 choices for the second, or 12*11 = 12!/(10!)=12P2. We conclude that the answer for 1 card is 12P1.
(2)
Fast way (not recommended).
Evaluate the answers numerically, and you can eliminate quite a few of them.

Answer 2

So it's A?

Answer 3

By the way, in my explanation above, I should have written
12P1=12!/(11!1!)=12
12C1=12!/(11!1!)=12
so 12P1=12C1

Answer 4

A says 12P1 so equals 12, does that sound right to you?

Answer 5

yes

Answer 6

Hmmm
You have 12 face cards, and 40 non-face cards.
Are there really only 12 ways to arrange the whole deck of 52 cards with a face card on top?
Think of it this way, there are 12 different face cards, so there are 12 ways to arrange the first card. There are 51 remaining cards to arrange...all in one single way?
In fact, there are n! ways to arrange n distinct objects, because if we choose the first out of n, then there are n-1 choices left for the next, n-2 the the still next until the last there is only one choice. So there are n! arrangements (permutations).
If there are 51 cards left after the first, then there are 51! possible arrangements, or permutations, which is equal to 51P51 (51 pick 51)

Answer 7

tell me the answer

Answer 8

I think its C