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Assume that there are 6 color letters L1, L2, L3, L4, L5 and L6 are to be placed in 6 same color envelop E1, E2, E3, E4, E5 and E6 (one letter for each envelop). The number of possibilities to place exactly one letter in its correct envelop ________.
in Combinatory by Active (2k points) 258 views
+3
$\binom{6}{1}$ * ( derangement of 5 element) = 6 * 44 = 264.
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if same Q ask for exactly  two letter in correct position then

15*9=135 ??   right
+1
yes, correct answer given by @Hemant. select 1 letter out of 6, put it in the correct envelope, and find derangement for 5 letters.

D5 = 44, hence answer will be 6*44.

$Remark:$ D5 = 5!( 1/0! - 1/1! + 1/2! - 1/3! + 1/4! -1/5!) = 44
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Can you please tell me from where to learn the concept of derangement. Any good source?
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@rohit, keneath rosen Discrete Maths book
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Thanks Manu Thakur.
+1
Number of derangement $\approx$$\frac{n!}{e} $ = $ \frac{5!}{e} =  264.87$

will work for multiple choices :p if stucked

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