Arranging the Cards into the Envelopes!

Probability Level 3

Six Cards and six envelopes are numbered 1,2,3,4,5,6. These cards are to be placed in envelopes so that each envelope contains exactly one card and no card is placed in the envelope bearing the same number. Moreover Card 1 is placed in Envelope 2. Then the number of ways it can be done is?


The answer is 53.

Ganesh Iyer by Ganesh Iyer , 22, India

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2 solutions

Tran Quoc Dat
Apr 22, 2016

Relevant wiki: Derangements

The number of ways the cards be placed (we now doesn't care about card 1 in envelope 2) is ! 6 = 265 !6 = 265 .

Now, among these ways, the number of ways card 1 is placed in envelope 2 equals to that when card 1 is placed in envelope 3, 4, 5 or 6.

So the answer is 1 5 × 265 = 53 \dfrac15 \times 265 =\boxed{53} .

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Can you explain more clearly

Sid Rana - 4 years, 3 months ago
Ganesh Iyer
Apr 19, 2016

Using the principle of inclusion and exclusion,

the number of ways the cards can be placed is

=5! - [ ( 4 1 ) \binom{4}{1} 4! - ( 4 2 ) \binom{4}{2} 3! + ( 4 3 ) \binom{4}{3} 2! - ( 4 4 ) \binom{4}{4} 1!]

=120 - [96-36+8-1]

=53

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