Read up on derangements.
Q- There are 5 envelopes corresponding to the 5 letters.If the letters are placed in the envelopes at random. Find the probability that no single letter is placed in right envelope ?
Any help would be greatly appreciated.
Thanks,
Ashish
Read up on derangements.
Thanks a lot Plato,
I solved the above question and answer comes out to be
5!(1-1/1!+1/2!-1/3!+1/4!-1/5!) = 44.
There is one more similar problem.
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Four persons go to a birthday party. They leave their top-coats and hats in the lounge and pick them while returning back.The number of ways in which none of them picks up his own top-coat as well as his own hat is p. The number of ways in which exactly one of them picks up his own top-coat as well as his own hat is q. The number of ways in which a person picks up someone else’s top-coat and yet someone else’s hat is r. Then p+q+r is
(a) 117 (b) 104 (c) 113 (d) 108 (e) none of the foregoing
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I tried solving the above as mentioned below -
Sol - The number of ways in which none of them picks up his own top-coat as well as his own hat= no: of dearrangements one can form between the person and coat and person and hat = [4!*{1-1/1!+1/2!-1/3!+1/4!}] * [4!*{1-1/1!+1/2!-1/3!+1/4!}] = 81.
=> p = 81.
The number of ways in which exactly one of them picks up his own top-coat as well as his own hat = 4C1* [ 3!*{1-1/1!+1/2!-1/3!}] * 4C1 [ 3!*{1-1/1!+1/2!-1/3!}] = 64
=> q = 64.
The number of ways in which exactly one of them picks up his own top-coat as well as his own hat = 4C1* [ 3!*{1-1/1!+1/2!-1/3!}] * [ 3!*{1-1/1!+1/2!-1/3!}] = 16.
=> q = 16.
hence p+q+r = 81 + 64 + 18 = 163. hence option (e)
But the book answer is - 108
Please correct me where I am wrong.
Any help would be greatly appreciated.
Thanks,
Ashish