Probability of 3 aces from drawing 7 cards. 1.58 Intro to Probability, 2nd Ed.

Question

This is Problem 1.58 from Tsitsiklis, Bertsekas, Introduction to Probability, 2nd edition.

We draw the top 7 cards from a well-shuffled standard 52-card deck. Find the probability that: a) The 7 cards include exactly 3 aces. b) The 7 cards include exactly 2 kings. c) The probability that the 7 cards include exactly 3 aces, or exactly 2 kings, or both

My solution was

a) $${3! {49 \choose 4} \over {52 \choose 7} }$$

b) $${2! {50 \choose 5} \over {52 \choose 7} }$$

c) $${3! {49 \choose 4} \over {52 \choose 7} } + {2! {50 \choose 5} \over {52 \choose 7} } - {3! 2! {47 \choose 2} \over {52 \choose 7} }$$

The solution book says

book solution

Why do they ${4 \choose 3}$ instead of (3!)?

We have 7 slots where we can put the cards with each slot accepting 1 card. We want to fill the first 3 with aces and the last 4 with non-aces. There are 3! ways to distribute aces into the first 3 slots.

And then there are ${49 \choose 4}$ ways to fill the remaining slots.

There is another card problem where they use that same logic to solve the problem: question, answer

Answer 1

a) Divide all cards into two groups: a group of $4$ aces and a group of $48$ non-ace cards. From the first group you select any $3$ aces and from the second group you select any $4$ cards.

b) Again divide all cards into two groups: a group of $4$ kings and a group of $48$ non-king cards. From the first group you select any $2$ kings and from the second group you select any $5$ cards.

c) Use inclusion-exclusion principle. Adding a) and b) will double count the outcomes of exactly 3 aces and 2 kings, which must be subtracted once. To find the outcomes of exactly 3 aces and 2 kings, you divide all cards into $3$ groups: a group of $4$ aces, a group of $4$ kings and a group of $44$ non-ace and non-king cards. From the first group you select any $3$ aces, from the second you select any $2$ kings and from the last group you select any $2$ cards.

Answer 2

You're not filling "the first three slots" with the aces, you're just deciding which three aces are among the first seven cards, and which four or the remaining $48$ non-aces fill out the first seven. Note that it's $48$, not $49$ as in your solution, since you can't allow the fourth ace.