Six cards and six envelopes are numbered (1,2,3,4), 5,6 and 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 and moreover the card numbered 1 is always placed in envelope numbered 2 . Then the number of ways it can be done is: Permutations and Combinations (Mathematics)

(c) : Using the principle of inclusion and exclusion, we have the number of ways in which card number 1 be placed in envelope number 2

\begin{aligned} & =5!-\left\{{ }^4 C_1 \cdot 4!-{ }^4 C_2 \cdot 3!+{ }^4 C_3 \cdot 2!-{ }^4 C_4 \cdot 1!\right\} \\ & =120-\{96-36+8-1\} \\ & =120-\{60+8-1\}=53 \end{aligned}

Alternative solution :

\begin{aligned} & \text { Fact }: D_n=n!\left[1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}+\cdots \cdot(-1)^n \frac{1}{n!}\right] \\ & \therefore D_5=5!\left[1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}\right]=44 \\ & D_4=4!\left[1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}\right]=9 \end{aligned}

The number of ways

=D_5+D_4=44+9=53

where

D_n

is the number of dearrangements of

n

objects.

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