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Find the probability of drawing an Ace of Heart, two cards of Number 7 and two Face Cards; Without replacement and In any order!

I knew the correct answer is (use combination formula nCr)

$$ P(E)= \frac{\#E}{\#S} = \frac{1C1\cdot4C2\cdot12C2}{52C5} = \frac{1\cdot6\cdot66}{2598960} =\frac{33}{216580}= 0.0001524 $$

I try an alternative method based on similar approach by Cyllinidra from one of the question in stackexchange.

The Probability of getting a Ace of Heart and then two number 7 cards in order and then two Face cards in order is

$$P(E) =\frac{1\cdot4\cdot3\cdot12\cdot11}{52\cdot51\cdot50\cdot49\cdot48} ~~~~~\text{(assume the order matters!)}$$

As the five cards are selected In Any Order, there are $nPn=5P5=5!=120 $ orders that give me an Ace of Heart, two number 7 cards and two face cards. Typical sample points: $$\{(AH, 7a, 7b, Fa, Fb), (AH, 7a, 7b, Fb, Fa), (AH, 7b, 7a, Fa, Fb), (AH, 7b, 7a, Fb, Fa), \dots\} $$
So, I take the original probability and multiply by 120 $$ P(E) =\left[\frac{1\cdot4\cdot3\cdot12\cdot11}{52\cdot51\cdot50\cdot49\cdot48}\right]\cdot120 = \frac{33}{54145} = 0.00060947. $$ This answer is exactly 4 times greater than the correct answer. I think the $nPn=5P5=5!=120$ is incorrect! But I cannot explain myself.

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    You count hands containing $7\spadesuit, 7\clubsuit$ and $7\clubsuit,7\spadesuit$ as if they were different, but they aren't. That's one factor of $2$. Similarly, you double count the pairs $F_1,F_2$ where $F_i$ denotes a face card. That's the other factor of $2$. – 2017-02-06

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The first is okay.   The probability of selecting 1 from 1 ace of hearts, 2 from 4 sevens, and 2 from 12 faces, when selecting any 5 from 52 cards without bias is: $$\newcommand{\ch}[2]{{^{#1}\mathrm C_{#2}}}\mathsf P(E) =\dfrac{\ch 1 1\ch 42 \ch{12}2}{\ch {52}5}$$

The second over counts because you have already counted the two ways to arrange sevens and two ways to arrange the faces in your listing $\rm (AH, 7a, 7b, Fc, Fd)$, as you made no restriction such that $a

Rather than looking at it as ways to arrange 5 cards, count ways to select places for the cards you select: one place for the ace, two places for the sevens, and two places for the faces.

$$\mathsf P(E) =\frac{1\cdot4\cdot3\cdot12\cdot11}{52\cdot51\cdot50\cdot49\cdot48} \cdot\ch 5{1,2,2}=\frac{1\cdot4\cdot3\cdot12\cdot11}{52\cdot51\cdot50\cdot49\cdot48} \cdot\dfrac{5!}{2!^2} = \dfrac{\ch 1 1\ch 42 \ch{12}2}{\ch {52}5}$$

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    I got it. I have to consider permutations with repetitions of indistinguishable objects, which are 2 sevens and 2 faces. Thank you so much! – 2017-02-07