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I am trying to solve the following problem and the answer I am getting is (6 - 1) = 5 but the answer is given as 2.

Problem

A father purchased dress for his 3 daughters. The dresses are of same color but different size and they are kept in dark room. In how many ways all the 3 daughters will not choose their own dress?

Question

Is my logic in below solution missing something or may be my solution is correct but the answer given in book is wrong?

My Solution

Total number of ways of selecting dresses = number of ways of selecting wrong sizes + number of ways of selecting the correct sizes

Using symbols, the above equation becomes t = w + c

We need to determine w, which we can by knowing t and c. t= 3! = 6 c = 1 since there is only one correct way of each daughter selecting the correct size.

Therefore, 6 = w + 1

so, w = 5

So, there are 5 ways in which the daughters can select the wrong dress according to my logic. BUT the answer given is 2.

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    Hint: There are only 2 ways that _all 3_ daughters can get the wrong dress. In some of your solutions 1 daughter gets the right dress & her sisters get each other's dress. Please see https://en.wikipedia.org/wiki/Derangement – 2017-01-24
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    Ok. I get it. In my solution w includes where 1 or 2 daughters get the correct dress but 2 or 1 get the wrong dress. Thanks for that – 2017-01-24
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    You still getting it wrong. If 2 girls get correct dress. Then its obvious that 3rd girl also gets correct. Subtract cases with 1 girl gets correct. – 2017-01-24

2 Answers 2

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Hint -

You are skipping the cases one daughter select right dress.

Subtract that also.

So we have 5 - 3 = 2

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    If any doubt please ask. – 2017-01-24
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Let the daughters be $A,B,C$ their right dresses be $d_A,d_B,d_C$

If $A$ selects $d_B,B$ has to select $d_C,C$ has to $C_A$

If $A$ selects $d_C,B$ has to select $d_A,C$ has to $C_B$

Possible dress selections :

$$\{A,d_A;B,d_B;C,d_C\}$$

$$\{A,d_A;B,d_C;C,d_B\}$$

$$\{A,d_B;B,d_C;C,d_A\}$$

$$\{A,d_B;B,d_A;C,d_C\}$$

$$\{A,d_C;B,d_A;C,d_B\}$$

$$\{A,d_C;B,d_B;C,d_A\}$$

The third & fifth are all wrong selections