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:
$\left( a \right)264$
$\left( b \right)265$
$\left( c \right)53$
$\left( d \right)67$

Answer 1

Hint: In this particular question use the concept of derangements of numbers of there are n different objects and we have to place these objects into n different boxes such that no object is placed into same number of box which is given as, $\left( {n!\left( {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}} - \dfrac{1}{{5!}} + ........ + {{\left( { - 1} \right)}^n}\dfrac{1}{{n!}}} \right)} \right)$ so use these concepts to reach the solution of the question.


Complete step by step answer:
Given data:
There are six cards and six envelopes are numbered 1, 2, 3, 4, 5, 6.
Now we have to place these cards into envelopes such that no card is placed into the same number of envelopes.
Now as we know that if there are n different objects and we have to place these objects into n different boxes such that no object is placed into same number of box which is given as, $\left( {n!\left( {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}} - \dfrac{1}{{5!}} + ........ + {{\left( { - 1} \right)}^n}\dfrac{1}{{n!}}} \right)} \right)$.
So in the given problem, n = 6 so we have,
Total number of derangements = $\left( {6!\left( {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}} - \dfrac{1}{{5!}} + ........ + {{\left( { - 1} \right)}^6}\dfrac{1}{{6!}}} \right)} \right)$
$ \Rightarrow \left( {6!\left( {1 - \dfrac{1}{{1!}} + \dfrac{1}{{2!}} - \dfrac{1}{{3!}} + \dfrac{1}{{4!}} - \dfrac{1}{{5!}} + \dfrac{1}{{6!}}} \right)} \right)$
Now as we that n! = n (n – 1) (n – 2) (n – 3)..... so we have,
$ \Rightarrow 6! - \dfrac{{6!}}{{1!}} + \dfrac{{6!}}{{2!}} - \dfrac{{6!}}{{3!}} + \dfrac{{6!}}{{4!}} - \dfrac{{6!}}{{5!}} + \dfrac{{6!}}{{6!}}$
$ \Rightarrow 6! - 6! + \dfrac{{6.5.4.3.2!}}{{2!}} - \dfrac{{6.5.4.3!}}{{3!}} + \dfrac{{6.5.4!}}{{4!}} - \dfrac{{6.5!}}{{5!}} + \dfrac{{6!}}{{6!}}$
$ \Rightarrow 360 - 120 + 30 - 6 + 1 = 265$
Now it is given that the card numbered 1 is always placed in envelope numbered 2.
So there are 5 possibilities of wrong placing of the card numbered 1.
So the total number of ways such that card numbered 1 is always placed in envelope numbered 2 and rest of the cards are not placed in the same numbered envelope = $\dfrac{{{\text{total number of ways}}}}{5}$
$ \Rightarrow \dfrac{{265}}{5} = 53$ Ways.
So this is the required answer.
Hence option (c) is the correct answer.

Note:
Whenever we face such types of questions the key concept we have to remember is that always recall the formula of derangement as well as n! which is all stated above, so simplify substitute in place of n, 6 as the total number of envelopes as well as the cards are 6 and simplify then divide by 5 according to given condition we will get the required answer.