Permutations and Combinations

Given that there are 15 identical gold coins to be distributed among 3 persons.
Also, each person gets at least 3 coins ⇒ total =3×3=9 coins
∴9 coins are already distributed in 1 ways. we now have 6 coins to be distributed between 3 people.
⇒x1+x2+x3=6,xi≥0
Number of
Number of solutions is given by
‌n+r−1Cr−1
‌⇒n=6,r=3
∴ Reanired solution
∴ Required Number =‌6+3−1C3−1
=‌8C2=28