Combinations - selecting 7 persons

[rajatgl16]

In how many ways 7 persons can be selected from 5 indian, 4 british and 2 chinise, if atleast 2 are to be selected from each country.

 

[rpf_rr]

you have to choose two chinese forced, at least two british and two indians, and one between british or indian. so you have

1*$$\frac{4!}{2!2!}$$*$$\frac{4!}{2!2!}$$*(3+2)=180

the last term: 3 indians+2 british

 

[rajatgl16]

hey I have answer booklet (not solution). And in it answer given is "100".

I tried it as,
At least 2 persons have to be slected form each country so :

ways of selecting 2 persons from 5 indians is ⁵C₂

ways of selecting 2 persons from 4 british is ⁴C₂

ways of selecting 2 persons from 2 chinese is ²C₂

Thus ways of slecting 6 persons form entire group is ⁵C₂ * ⁴C₂ * ²C₂

Now 1 person has to be selected from remaining 3 Indians and 2 british and 0 chinese

possible way of selceting 1 person is ⁵C₁

Thus final answer to select 7 persons is ⁵C₂ * ⁴C₂* ²C₂ * ⁵C₁=300 so its also wrong

 

[rpf_rr]

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.

 

[rajatgl16]

Then may be ans in my ans booklet is wrong.

 

[Office_Shredder]

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

 

[rpf_rr]

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

 

[rajatgl16]

Hmm. I was wrong. Thanks guys for helping me,