Derangements

Question

How many ways we can put 5 letters L1, L2, L3, L4, L5 in 5 envelopes e1, e2, e3, e4 and e5 (at 1 letter per envelope) so that

i. no letter is correctly placed?

ii. at least 1 letter is correctly placed?

iii. exactly 2 letters are correctly placed?

iv. at most 1 letter is correctly placed?

v. at least 1 letter is wrongly placed?

vi. exactly 1 letter is wrongly placed?

Answer 1

Derangement :- permutation of elements such that no element get place to its original position.

Standard Example :- There are 5 letters and 5 envelopes, place these letters in such a way that no letter is placed in right envelope so that it can be delivered to its right address.

1) put n=5 and calculate , you will get 44

2)at least one placed correctly =Total ways (5!)- no one placed correctly . (120-44=76)

3) Exactly 2 letters placed correctly :- first place those(way to choose any 2 out of 5 ,5C2=10) at their correct position (1 way) then do derangement with  remaining 3 letters (2 ways ) so ultimately 10*2=20

4) At most one letter is correctly placed :- No one correctly placed + exactly one correctly placed (44+9=53)

5) at least one wrongly placed :- Total ways to arrange - no one wrongly placed (5!-1=119)

6) exactly one is wrongly placed :- It cant possible as one is wrongly placed mean it occupied someone else's position so in that way that element is forced to place some where else. so here number of ways 0

for further interest , you can look here.

http://www.askiitians.com/iit-jee-algebra/permutations-and-combinations/derangements-and-multinomial-theorem.aspx

https://en.wikipedia.org/wiki/Derangement

Comments

Yes 89 is correct

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@Himanshu Kashyap

@Kaluti

yes 89 is correct

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at most one correctly placed is = no one correctly placed + exactly one correctly placed = D5 + 5C1 * D4=89