Number of ways 6 things can be selected from a group of identical things In how many ways can a party of 6 men be selected out of 10 Hindus, 8 Muslims, and 6 Christians if the party consists of at least one person of each religion( Consider only the religion of the person?


Now, I know this problem can be solved by finding the non-negative integral solution to the given equation,

$$x+y+z=6$$
Which the answer turns out to be $\binom{6-1}{3-1}$

But I want to know is there any other method of solving this problem? 

Any help would be appreciated.
No.of possibilities according to the given condition is as follows,
$$(1,1,4), (1,2,3), (2,2,2)$$

Now we find the no.of ways we can select the persons of different religion corresponding to the possibilities.

1) For $(1,1,4)$,

No.of ways of selecting the persons = $3!$. 

But since there are 2 1's, each selection will have two copies. 

And since we only need to consider their religion (that means persons in a particular religion are identical), we need to divide the result by 2. That gives us,

$\frac{3!}{2} = 3$ ways

For $(1,2,3)$

No.of ways of selecting the persons = $3!$.

(Notice there aren't any repeating numbers in this case!)

2) For $(2,2,2)$

No.of ways of selecting the persons = $3!$.

Now, there are 3 repeating numbers.

That means there are $3!$ copies for each selection. Applying the same logic, we divide the result by $3!$.

$\frac{3!}{3!} = 1$

Adding all the result gives us,

$3+3!+1 = 10$
The number of ways arranging this would be 10 . 

And the possibilities are :

( 1 , 1 , 4)
( 1 , 2 , 3)
( 1 , 3 , 2)
( 2 , 2 , 2)
( 2 , 3 , 1)
( 2 , 1 , 3)
( 3 , 2 , 1)
( 3 , 1 , 2)
( 4 , 1 , 1)
( 1, 4 , 1)

following the pattern ( Hindu, Muslim , Christian )