Question

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 the card numbered 1 is always placed in envelope numbered 2. Then the number of ways it
can it be done?
A .264
B .265
C .53
D .67

Answer 1

Hint: Use derangement formula,
\[{{D}_{n}}=n!\left( 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}...........+{{(-1)}^{n}}\frac{1}{n!}

\right)\]

Envelope: 1 2 3 4 5 6

Card: 1 2 3 4 5 6

Complete step-by-step answer:

According to the question, it is given that card no 1 is placed in envelope number 2.

Envelope: 2 1 3 4 5 6

Card: 1 2 3 4 5 6

Assume card no.2 is present in an envelope no. 1.

$2\to 1$

Now we have the following situation,

Envelope: 3 4 5 6

Card: 3 4 5 6

Cards should be placed in envelopes such that the envelope and card don’t have the same number.

It means we have to dearrange four cards and envelopes.

The formula, \[{{D}_{n}}=n!\left(

1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}...........+{{(-1)}^{n}}\frac{1}{n!} \right)\] is to be used.

\[\begin{align}

  & {{D}_{4}}=4!\left( 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!} \right) \\

 & \Rightarrow {{D}_{4}}=9 \\

\end{align}\]


If card 2 is not present in envelope 1, then we have to rearrange 5.

Derangement formula to be used,

\[\begin{align}

  & {{D}_{5}}=5!\left( 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{4!}-\frac{1}{5!} \right) \\
 & \Rightarrow {{D}_{5}}=44 \\

\end{align}\]

Total possible ways=44+9

                                  =53.

Note: Remember derangement formula,\[{{D}_{n}}=n!\left( 1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}...........+{{(-1)}^{n}}\frac{1}{n!} \right)\] The formula is especially useful in cases where letters are to be placed in envelopes or balls are to be placed in boxes.