Exactly 3

Probability Level 3

8 balls are laid out in a 2 rows.

How many ways can they be rearranged so that exactly 3 are in the same position?

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1 solution

Geoff Pilling
Nov 10, 2018

There are $\binom{8}{3} = 56$ ways of choosing which $3$ to be in the same position.

And there are $!5 = 44$ ways of rearranging the remaining $5$ , where $!n$ is the derangement formula.

$56 \cdot 44 = \boxed{2464}$

A non-derangement variation: How many ways can they be rearranged so that exactly 4 are in the same row that they started in? I'm getting $20736$ .

Brian Charlesworth - 2 years, 7 months ago

I get the same.

6 ways to choose 2 from row 1 to move to row 2. 6 ways to choose 2 from row 2 to move to row 1. Then 24 ways to rearrange each of the rows.

$6 \cdot 6 \cdot 24 \cdot 24=20736$

Geoff Pilling - 2 years, 7 months ago

Yeah sir.......this is the same approach I used..........!!!

Aaghaz Mahajan - 2 years, 7 months ago