for nCk you mean n!/(k!*(n-k)!) ? If yes we computed the same thing, or better, you are right, i've done an error in the third factor, it is actually 5C2=5!/(3!2!), for me it's 600, but i made the same reasoning as you did, and i think it's right, if we understand the problem correctly.
You've overcounted some. Imagine the British people are labeled A,B,C and D.
Scenario 1: You pick two British, A and B. Then you pick two Indians. Then you pick your last person from the five remaining people and pick person C.
Now imagine instead you pick two British, A and C. You pick the same two Indians as before. You pick your last person from the five remaining people and the person is B.
In both situations you've picked the same set of people but you counted them separately
Officeshredder is right. There are two possible situations, you pick 2 chinese 3 british and 2 indians or you pick 2 chinese 2 british and 3 indians, so you have
2C2*4C3*5C2+2C2*4C2*5C3=100