Number of ways 6 things can be selected from a group of identical things

Question

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.

Answer 1

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 )

Answer 2

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,

$3+3!+1 = 10$

Answer 3

Stars and Bars

Once we pre-select one of each religion, this is the same as asking "how many ways we can choose $3$ men from $9$ Hindus, $7$ Muslims, and $5$ Christians?" This is the same as asking how many ways to solve $$ x+y+z=3 $$ with $x,y,z\ge0$. Stars and Bars say $\binom{5}{2}=10$.

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Generating Functions

$$ \begin{align} &\left[x^n\right] \overbrace{\left(x+\cdots+x^{10}\right)}^{1\dots10\text{ Hindus}} \overbrace{\left(x+\cdots+x^8\right)}^{1\dots8\text{ Muslims}} \overbrace{\left(x+\cdots+x^6\right)}^{1\dots6\text{ Christians}}\\ &=\left[x^n\right]x^3\frac{1-x^{10}}{1-x}\frac{1-x^8}{1-x}\frac{1-x^6}{1-x}\\ &=\left[x^{n-3}\right]\frac{1-x^6-x^8-x^{10}+x^{14}+x^{16}+x^{18}-x^{24}}{(1-x)^3}\\ &=\left(\left[x^{n-3}\right]-\left[x^{n-9}\right]-\left[x^{n-11}\right]-\left[x^{n-13}\right]+\left[x^{n-17}\right]+\left[x^{n-19}\right]+\left[x^{n-21}\right]-\left[x^{n-27}\right]\right)\frac1{(1-x)^3}\\ &=\textstyle(-1)^{n-3}\left[\binom{-3}{n-3}-\binom{-3}{n-9}-\binom{-3}{n-11}-\binom{-3}{n-13}+\binom{-3}{n-17}+\binom{-3}{n-19}+\binom{-3}{n-21}-\binom{-3}{n-27}\right]\\[6pt] &=\bbox[5px,border:2px solid #C0A000]{\textstyle\binom{n-1}{n-3}-\binom{n-7}{n-9}-\binom{n-9}{n-11}-\binom{n-11}{n-13}+\binom{n-15}{n-17}+\binom{n-17}{n-19}+\binom{n-19}{n-21}-\binom{n-25}{n-27}} \end{align} $$ This is the number of ways to pick a group of $n$ people from a pool of $10$ Hindus, $8$ Muslims, and $6$ Christians, where there is at least $1$ of each.

The only one of these that is not $0$ for $n=6$ is $\binom{n-1}{n-3}=\binom{5}{3}$.

Note that we could reduce the last line to $$ \textstyle\binom{n-1}{2}-\binom{n-7}{2}-\binom{n-9}{2}-\binom{n-11}{2}+\binom{n-15}{2}+\binom{n-17}{2}+\binom{n-19}{2}-\binom{n-25}{2} $$ except that the equation $\binom{n}{k}=\binom{n}{n-k}$ only holds when $k$ and $n-k$ are non-negative integers.