Posting a letter wrong!

Probability Level 3

There are n n envelopes and n n corresponding letters. Find the number of ways in which the envelopes can be distributed such that exactly one letter goes in the wrong envelope in terms of n n .


Notations:

  • C ( a , b ) C(a,b) is the total number of ways way of choosing b b elements from a set of a a elements in which order does NOT matter.

  • P ( a , b ) P(a,b) is the total number of ways way of choosing b b elements from a set of a a elements in which order does matter.

P ( n , n − 1 ) P(n,n-1) n ! ( 1 2 ! − 1 3 ! + ⋯ + ( − 1 ) n ⋅ 1 n ! ) n! \left( \frac1{2!} - \frac1{3!} + \cdots + (-1)^n \cdot \frac1{n!} \right) C ( n , n ) − 1 C(n,n) -1 n ! ( 1 2 ! − 1 3 ! + ⋯ ) n! \left( \frac1{2!} - \frac1{3!} + \cdots \right)

Daman Deep Singh by Daman Deep singh , 22, India

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1 solution

Daman Deep Singh
Jan 25, 2017

If you put ( n − 1 ) (n-1) letters in their corresponding ( n − 1 ) (n-1) envelopes then you are only left with one letter and one envelope which makes the right pair. So, there is N O NO such way of putting exactly one letter in wrong envelope. :)

2 Helpful 0 Interesting 0 Brilliant 0 Confused

While setting the options, I intended that. ;)

Daman Deep singh - 4 years, 4 months ago

Haha I was thinking of filing a report saying "But, this is impossible!" until I noticed that one of the answer options was indeed 0 0 . :)

The number of ways of distributing the letters so that precisely one of them is c o r r e c t l y correctly placed is of course n D ( n − 1 ) nD(n-1) , where D ( k ) D(k) is the number of derangements of k k distinct objects.

Brian Charlesworth - 4 years, 4 months ago

0 pending reports

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