Discussion Forum

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

Answer:

Explanation:

 Plan  If there are n different objects  and n corresponding places, then the number of ways of  putting these  object to  the no object occupies its corresponding place

                   = $n!\left[\frac{1}{2!}-\frac{1}{3!}+....(-1)^{n}\frac{1}{n!}\right]$

 We have six cards C₁,C₂,C₃,C4,C₅,C₆

 and envelope E₁,E₂,E_3,E₄,E₅,E₆

Let the number of rearrangement when C₁  is put into E₂ =X

Similarly, the number of rearrangement when C₁  is put into E₃=X

Similarly, the number of rearrangement when C₁  is put into E₄=X

Similarly, the number of rearrangement when C₁  is put into E₅=X

Similarly, the number of rearrangement when C₁  is put into E₆=X

 Thus, total number of rearrangement =5X=D₆

 D₆= $6!\left[\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!}+\frac{1}{6!}\right]=265$

$\therefore$     $  X=265/5=53$