This Can Only Happen in India!

It's a question using the concept of de-arrangement .

Formula for de-arrangement of exactly $r$ objects out of total $n$ objects is : $\binom{n}{r} \times (r!) \times (\dfrac{1}{2!}-\dfrac{1}{3!}+\dfrac{1}{4!}-......+(-1)^{r+1}\dfrac{1}{r!})$

Here, let "case $\lambda$ " means de-arrangement of $\lambda$ jackets (exactly) out of their total $7$ jackets.

$\begin{cases} case 0=1 \\ case 1 = 0 \\case 2 =21 \\case 3= 70 \\case 4=315 \end{cases}$

So de-arrangement of at the most $4$ of jackets $=1+0+21+70+315=\boxed{407}$

Definition of de-arrangement : De-arrangement means the objects are not at their actual (previous or where they should be) positions. In this case de-arrangement represents that they don't get their own jackets.

And $\binom{n}{r}=\dfrac{n!}{r! \times (n-r)!}$

I like Sandeep's solution, however, I didn't remember the formula's for de-arrangement's off the top of my head. My solution uses the idea of distoint cycles in permutations, a topic usually covered in abstract algebra.

So for a permutation on $7$ objects, $r$ people not getting their jackets means that the sum of the lengths of the non-trivial cycles is $r$ (since a trivial cycle [a cycle of size $1$ ] means the person got their own jacket).

For $4$ jackets, we can either have $2$ disjoint cycles of size $2$ or $1$ disjoint cycle of size $4$ . For $3$ jackets, the only possibility is one disjoint cycle of size $3$ . For $2$ jackets, also the only possibility is one disjoint cycle of size $2$ . It's impossible for only one person to not get their own jacket. And lastly, let's not forget the single case in which everyone gets their own jacket.

Case 1: $2$ disjoint cycles of size $2$

For the first disjoint cycle, we have $_7P_2$ options. For the second disjoint cycle, we've already selected $2$ , leaving $5$ left, so we have $_5P_2$ options. For each cycle, we are overcounting by a factor of $2$ because we want the number of permutations on a circle (in this case, a circle of size $2$ ). In addition, we don't care about the order of which we pick the two cycles either, so we are also overcounting by a factor of $2$ there. So our final formula is:

$(\frac{_7P_2}{2}\frac{_5P_2}{2})/2 = 105$

Case 2: $1$ disjoint cycles of size $4$

So there are $_7P_4$ ways to pick $4$ jackets. We are overcounting by a factor of $4$ since we want permutations on a circle of size $4$ . We only have a single cycle, so there is no overcounting in terms of permutations of the cycles themselves.

$\frac{_7P_4}{4} = 210$

Case 3 and 4: $1$ disjoint cycle of size $3$ , $1$ disjoint cycle of size $2$

Using the same reasoning as above, the formulas are:

$\frac{_7P_3}{3} = 70$

and

$\frac{_7P_2}{2} = 21$

respectively.

Case 5: Everyone gets their own jacket. There is only one possible way this can happen.

Summing up all the possible cases: $105+210+70+21+1 = 407$