JEE (Advanced) 2014 Combinatorics

• Quick way, (i.e., by cheating)

The number of derangements of a set of $6$ items is $265$ .

One fifth of these will involve card $1$ being placed in envelope $2$ , giving us $(\frac{1}{5}) * 265 = \boxed{53}$ permutations under the given conditions.

• Long way: We need to look at some cases:

--- if card $2$ is placed in envelope $1$ then we are left with a derangement of $4$ items, namely $3, 4, 5, 6$ . There are $9$ such permutations.

--- if card $2$ is placed in one of envelopes $3, 4, 5, 6$ , say envelope $n$ , and then card $n$ is then placed in envelope $1$ then we've formed a cycle, leaving us with $3$ remaining items to derange in $2$ possible ways. This gives us $4*2 = 8$ permutations.

--- if card $2$ is placed in one of envelopes $3, 4, 5, 6$ , say envelope $n$ , and then a card other than card $n$ is placed in envelope $1$ , then card $n$ has $3$ remaining envelopes in which it can be placed. The remaining two cards will then have just one way in which they can be placed in envelopes as required. This gives us a total of $4*3*3 = 36$ permutations.

This gives us a total of $9 + 8 + 36 = \boxed{53}$ permutations as found before.

Method 1: Using Inclusion exclusion principle

Total number of arrangements= $5!$

We need to subtract the cases when cards are in right places given that Card numbered $1$ is always placed in envelope $2$

Total number of ways= $5! - (4 \cdot 4! -{4 \choose 2} \cdot 3! + {4 \choose 3}\cdot 2! -1) =53$

Method 2: Dearrangement

If we place card $2$ in envelope $1$ then we need to dearrange $4$ things which can be done in $9$ ways (Because $D_{4}=9$ )

If card $2$ doesn't go in $1$ we need to dearrange $5$ things which can be done in $44$ ways.

Answer= $9+44=\boxed{53}$