How often do 4 teams of 7 get narrowed to 1 team of 5, 1 team of 1, and 2 teams of 0

Question

There are 4 teams of 7 randomly picked people each. At the end of each day, one person gets randomly eliminated. When there are 6 people remaining, what is the probability that one team has 5 people, one team has one person, and the other two teams have no remaining people? It doesn't matter which team, just that this 5-1-0-0 scenario happens(can be 0-5-1-0, 0-0-5-1, 1-0-5-0, etc.)

I think the denominator of this probability should be 28 choose 22 since that represents the number of ways we can select 22 people. However I am then unsure how to formulate the numerator because it does not matter which team has 5 people in it, just that one of them does.

Thanks

Answer 1

Take a standard deck of playing cards. Select out 7 cards of each suit, making a pack of 28 cards. Shuffle them. Deal out 6 cards.

What are the odds that you have 5 cards of one suit and 1 card of a different suit?

This is the same question you are asking. Hopefully framing it differently helps.

(In other words, the "random elimination" doesn't matter whatsoever. Just choose six random people.)

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The answer should be:

$${4 \choose 1}{7 \choose 5}{21 \choose 1} \over {28 \choose 6}$$

Choose one suit out of four to be the suit with 5 cards remaining; choose the specific 5 cards remaining; choose the 1 leftover card from the other three suits; divide by the total ways to make 6 cards from the pack of 28.