21Number of ways to place 4 letters in 4 envelopes without any condition = 4!
Number of ways to place all letters correctly into the corresponding envelopes = 1
Number of ways to place one letter is the wrong envelop and other 3 letters in the right envelope = 0
(Because it is not possible that only one letter goes in the wrong envelop)
Number of ways to place atleast two letters in the wrong envelopes
= Total number of way to place letters
= Number of ways to place all letters correctly
= Number of ways to place on letter correctly = 4! – 1 – 0 = 23
49no of way to put 1 letters in 1 wrng envelope=0
no of way to put 2 letters in 2 wrng envelope=1
no of way to put 3 letters in 3 wrong envelope=all permutation - wen 1 is correct - wen all 3 are correct=3!-3C1*1-1=2
P.S::2 correct means all correct since 2 correct will ensure the third will also be a correct one!
no of ways to put 4 in 4 incorrect envelopes=all-1correct-2correct-all correct=4!-4C1*2-4C2*1-1=24-8-6-1=9
check aswers!!
49
Subhomoy Bakshi
·2010-05-15 06:32:56
for number (i)
at least 2 in wrong=all permutation-1 in correct=4!-4C1*2=16
1
Sridhar M
·2010-05-15 06:43:34
Another way of doing this
No of derangements (wrong arrangements) Dn = n! [1- 11! + 12! + ........(-1)nn!]
we need D4 = 4! [12 - 16 + 124] = 9