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The question is in the title itself. First, I would like to share how I solved this problem at first:

We have $4$ coats, $5$ waistcoats and $6$ caps. So, I considered that each man wears one coat, one cap and one waistcoat at a time.

So, one cap, one coat and one waistcoat makes one dress.
$\therefore$ total dresses = $6 * 5 * 4 = 120$

Now, all we have to do is find the permutations of 120 dresses taken 3 at a time:
$120! / 117! = 120 * 119 * 118 = 1685040$

However , my book gives me a different solution:
Number of ways in which 3 men can wear 4 coats = $P(4 , 3) = 4!$
Number of ways in which 3 men can wear 5 waistcoats = $P(5,3) = 5 * 4 * 3$
Number of ways in which 3 men can wear 6 caps = $P(6,3) = 6 * 5 * 4$
So , total number of ways to wear these clothes = Product of all the above RHS terms which gives the answer 172800.

I agree with this solution too. But why am I getting a wrong answer?

Takahiro Waki
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codetalker
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2 Answers2

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Because many of your arrangements are mutually exclusive because they include the same unique article of clothing. If one man is wearing the big red hat, any combination using that article is invalid.

An alternate route to the answer more along the lines of your thinking would be to dress the first man $6\times5\times4$. Decrement each category by one. Dress the second man: $5\times4\times3$, then dress the third man: $4\times3\times2$. Multiplying those 3 terms will give you the correct answer.

turkeyhundt
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You are indeed overcounting. See if this helps:

Let $A,$ $B,$ and $C$ be the three men. There are charcoal, black, brown, and white suits available. With your count, $A,$ $B,$ and $C$ could all be wearing the same suit! To avoid this, we do casework based on the article of clothing to avoid this sharing of clothes.

K. Jiang
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