If two man have the same $ 3 \ $ pairs of shoes, the same $3 \ $ pairs of pants, the same $ 3 \ $ shirts, and the same $ \ 3$ sweaters. How many ways can they dress so that are not both dressed exactly the same?
Answer:
Suppose one man choose to wear all $ \ 4 $ items but the second man reject anyone item and choosen the rest three item , then they can be dress up not exactly same with
$$
\underbrace{(6 \times 6 \times 6 \times 6)}_{\text{for the first man}}
\times
\underbrace{(5 \times 5 \times 5)}_{\text{for the $2$nd man}}
$$
ways.
But since there are $4$ types of item , the first man can choose the fourth item in $4$ ways.
Thus the total ways to dress up the two mans is
$$
4
\times
\underbrace{(6 \times 6 \times 6 \times 6)}_{\text{for the first man}}
\times
\underbrace{(5 \times 5 \times 5)}_{\text{for the $2$nd man}}.
$$
I need help solving this problem.
Best Answer
I'm not sure at all where you're getting the $6$'s from in your answer, but here's a hint.
Calculate the number of ways that they can dress exactly the same, and subtract that number from the total number of ways they can both dress.
How many ways can one of the men dress?
Spoiler: