A typist has six envelopes and six letters.In how many ways can one letter be placed in each envelope with out getting every letter in the right envelope
This is a classic combinatorial problem called the *derangements* problem. A derangement is a permutation of the elements of a set such that none of the elements appears in its original position. In this case, there are 6 letters and 6 envelopes, so there are 6! (i.e., 6 factorial) total possible ways to place the letters in the envelopes. We are asked to find the number of these permutations that are derangements.
There is a formula for finding the number of derangements for a set of size n, called the subfactorial !n:
!n = n! * (1 - (1/1!) + (1/2!) - (1/3!) + ... + (-1)^n * (1/n!))
For n = 6, we have:
!6 = 6! * (1 - 1 + 1/2 - 1/6 + 1/24 - 1/120)
!6 = 720 * (1 - 1 + 0.5 - 1/6 + 1/24 - 1/120)
!6 = 720 * (0.5 - 0.1667 + 0.04167 - 0.008333)
!6 = 720 * (0.3667)
So, rounding to the nearest whole number, there are approximately 265 derangements. Therefore, there are 265 ways to place the letters in the envelopes without getting every letter in the right envelope.