Question

Seven different coins are to be divided amongst three persons. If no two of the persons receive the same number of coins but each receives atleast one coin & none is left over, then the number of ways in which the division made is k, then the sum of digits in number of k equals.

Answer 1

Well with three people and no two people getting the same number (and everyone gets atleast 1)

The only way is to divide is 1,2,&4

Assume person one gets one coin there are $\to$ 7 possibilities

Person two gets 2 coins from the remaining 6=

$\to \frac{6!}{(6-2)!2!}=15$

(because the order of the coins does not matter)

Person 3 gets what's left-4 coins (1 possibility again order does not matter)

but there also $\to 6$ ways to arrange the people

So,

$k=7\times 15\times 6=630$

Now ,sum of digits in number $k=9$