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 moreover the card numbered $1$ is always placed in envelope numbered $2$. Then, the number of ways it can be done is:
(A) $264$
(B) $265$
(C) $53$
(D) $67$

Answer 1

The given question revolves around the concepts of permutations and combinations. We have to place the six cards in six envelopes such that no card goes in the envelope with the same number and the card numbered $1$ placed in the envelope numbered $2$. The problem given to us can be solved easily using the derangement formula and related concepts.

Complete step by step solution:
In the given question, we have to calculate the number of ways in which no card goes in the envelope with the same number and the card numbered $1$ placed in envelope numbered $2$.
The derangement can be explained as the permutation of the elements of a certain set in a way that no element of that set appears in their original positions.
Therefore, the derangement formula can be applied in the given problem.
Derangement formula is $D\left( n \right) = n!\left[ {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}}.... + \dfrac{{{{\left( { - 1} \right)}^n}}}{{n!}}} \right]$.
So, the number of ways in which no card goes in the envelope with the same number in this case is $D\left( 6 \right)$. But out of these derangements, there are five ways in which a card numbered $1$ is going wrong.
So, when card numbered $1$ goes in envelope $2$, the number of ways are $\dfrac{{D\left( 6 \right)}}{5}$.
Now, we have to calculate the value of $D\left( 6 \right)$.
So, $D\left( 6 \right) = 6!\left[ {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}} - \dfrac{1}{{5!}} + \dfrac{1}{{6!}}} \right]$
Substituting the values of factorials, we get,
$\Rightarrow D\left( 6 \right) = 720\left[ {1 - \dfrac{1}{1} + \dfrac{1}{2} - \dfrac{1}{6} + \dfrac{1}{{24}} - \dfrac{1}{{120}} + \dfrac{1}{{720}}} \right]$
Taking LCM, we get,
$\Rightarrow D\left( 6 \right) = 720\left[ {\dfrac{{720 - 720 + 360 - 120 + 30 - 6 + 1}}{{720}}} \right]$
Simplifying further, we get,
$\Rightarrow D\left( 6 \right) = 360 - 120 + 30 - 6 + 1$
$\Rightarrow D\left( 6 \right) = 265$
Therefore, the number of ways in which no card goes in the envelope with the same number and the card numbered $1$ placed in envelope numbered $2$ is $\dfrac{{D\left( 6 \right)}}{5} = 53$. So, option (C) is correct.

Note:
Derangement can be simply defined as a permutation arrangement with no fixed points. One must know the concepts of derangement thoroughly so as to solve such questions. Care should be taken while calculating the values of factorials and derangements.