Latest Maths Olympiad Aptitude Question SOLUTION: Q. 6 gentlemen visit a club, and leave their hats with the hatrack lady. At the end of the event, the lady forgot who brought which hat, and distribute the hats randomly. How many

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Q. 6 gentlemen visit a club, and leave their hats with the hatrack lady. At the end of the event, the lady forgot who brought which hat, and distribute the hats randomly. How many ways are there for the lady to get at most 1 gentleman's hat correctly?

Asked In Maths Olympiad SDeep Bhattacharya (12 years ago)

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SDeep Bhattacharya  (12 years ago)

6 gentlemen visit a club, and leave their hats with the hatrack lady. At the end of the event, the lady forgot who brought which hat, and distribute the hats randomly. How many ways are there for the lady to get at most the hat of 1 gentleman correctly?

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Ram Prakash  (11 years ago)

Total number of ways of distributing 6 hats among the gentlemen = 6! = 720
Number of ways of distributing 6 hats such that everyone gets their respective hat = 1
Number of ways of distributing 6 hats such that exactly 4 of them get their respective hat 6C4*9=135
Number of ways of distributing 6 hats such that exactly 3 of them get their respective hat = 6C3*2 = 40
Number of ways of distributing 6 hats such that exactly 2 of them get their respective hat = 6C2*1=15
Therefore, number of ways of distributing hats such that at most 1 gentlemen gets correct hat = 720 – (135 + 40 + 15 +1) = 529

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Ram Prakash  (11 years ago)

Total number of ways of distributing 6 hats among the gentlemen = 6! = 720
Number of ways of distributing 6 hats such that everyone gets their respective hat = 1
Number of ways of distributing 6 hats such that exactly 4 of them get their respective hat 6C4*1=15
Number of ways of distributing 6 hats such that exactly 3 of them get their respective hat = 6C3*2 = 40
Number of ways of distributing 6 hats such that exactly 2 of them get their respective hat = 6C2*9=135
Therefore, number of ways of distributing hats such that at most 1 gentlemen gets correct hat = 720 - (1 + 15 + 40 + 135) = 529.
Note: Multiplication factor 9, 2 and 1 are basically the number of ways 4, 3 and 2 persons hat distributed among them such that nobody gets their respective hat

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Samik Mukherjee  (11 years ago)

The question asks at most 1 right delivery of hat. so there are two choices; no right choice or 1 right choice. out of total 720 choices (6!), only 6 are those where once gentleman gets his own hat while the others don't. and there is one choice where none of them get their own hat. so the total no. of ways=6+1=7.

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Narendra Kulkarni  (12 years ago)

We have to consider two cases, zero correct matching and one correct matching. There is only one case possible for zero correct matching and for one correct matching, there are 6 cases as any one out of 6 persons can get his hat correctly.
Hence total number of ways=1+6=7.
Thus, 7 is the answer.

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Rakesh  (12 years ago)

i think the answer is 11
considering none of them get their hat , we have cases(2 possiblesties for each macking it 2*2*2) how ever we can have atmost 1 correct hat, that gives us 3 combinations(1 correct hat for one person at a time), thta makes 8+3=11 combinations

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