STEP 2013 P1 Statistics and Probability Question The aim to the following: From a deck of 52 cards (numbered from 1,2,3,...,52) I pick 7 cards, each card having the same probability of being picked. What is the probability that only two of the selected cards add up to 53?
My approach:
Say we have the cards $A,B,C,D,E,F \text{ and } G$. The probability that a randomly selected card, say $A$, from these $7$ has its pair in the set is $\frac{6}{51}$. Say this pair is $B$. Now we need to figure out the number of other possible pairs and the probability that a given number doesn't have its pair. The probability that a number doesn't have its pair (from the $5$ numbers left) is $1-\frac{5}{51}=\frac{46}{51}$. There are $\binom{5}{2}=10$ many different possible pairs. Thus the probability that exactly one pair add up to $53$ is
$$p=\frac{6}{51}(\frac{46}{51})^{10}.$$
According to the markscheme this is wrong. Could somebody let me know why is that?
 A: There is a natural pairing of the cards into $26$ pairs that sum to $53$: $1-52,2-51$ and so on. Indeed, these are the only ways to make $53$ from two cards in the set.
The number of hands with only two cards – one pair – summing to $53$ is the product of

*

*$26$ ways to select the pair in hand

*$\binom{25}5$ ways to select the pairs the other five picked cards reside in – each of these cards must reside in its own pair

*$2^5=32$ ways to select which card in the selected single-card pairs is actually in the hand

Thus the final probability is
$$\frac{26×\binom{25}5×32}{\binom{52}7}$$
A: There are $52\cdot51\cdot50\cdots46$ ways to choose $7$ cards. There are $\binom72=21$ ways to choose which two cards will be paired, $26$ ways to choose the pair, and $2$ ways to distribute the pair to the chosen positions.
The first of the remaining $5$ cards can be chosen in $50$ ways, the next in $48$ ways, and so on, because once we choose a card we cannot choose it or its pair.  This gives a probability of $$\frac{21\cdot52\cdot50\cdot48\cdot46\cdot44\cdot42}{52\cdot51\cdot50\cdot49\cdot48\cdot47\cdot46}$$
