<s>[INST] sum which have binomial coefficients 
Finding value of
$\displaystyle \binom{50}{6}-\binom{5}{1}\binom{40}{6}+\binom{5}{2}\binom{30}{6}-\binom{5}{3}\binom{20}{6}+\binom{5}{4}\binom{10}{6}$

Try: Equation coefficient of $x^6$ on both side
$$\bigg[(1+x)^{10}-1\bigg]^5=(1+x)^{50}-\binom{5}{1}(1+x)^{40}+\binom{5}{2}(1+x)^{30}-\binom{5}{3}(1+x)^{20}+\binom{5}{4}(1+x)^{10}$$
So our required sum is coefficient of $x^6$ in $$x^5\cdot \bigg[1+(1+x)+(1+x)^2+\cdots +(1+x)^9\bigg]^5.$$
So coefficient of $x$ in $$\bigg[1+(1+x)+(1+x)^2+\cdots \cdots +(1+x)^{9}\bigg]^5$$
$$=5(10)^4 (1+2+\cdots +9)=5\cdot (10)^4\cdot 45$$
But I am  also trying to solve it using the principle of inclusion -exclusion.
I have seems that we will form $5$ groups and selecting some persons from each group, but I did not understand how I can solve it
Could some help me to solve it , thanks
 [/INST] Let's consider the following problem: Given 50 balls colored in five colors, 10 per color, how many ways can we choose 6 balls so that each color is represented?
Denote the answer $X$.
We will solve this problem in two different ways. Both should give the same result.
First, let's solve the above using inclusion-exclusion principle.


*

*We count how many ways we can choose 6 balls. That is $50 \choose 6$

*Now we remove all combinations that miss a color. How many are there? There are 5 possibilities for the missing color, and 40 balls colored differently. Thus, subtract ${5 \choose 1}{40 \choose 6}$ combinations.

*But arrangements missing two colors were counted twice in step 2, so we need to add them back once. There are $5 \choose 2$ ways to select two missing colors. This leaves 20 balls, so we have \choose 6$ ways to choose the balls.

*To complete the inclusion and exclusion, we have to subtract arrangements missing 3 colors, and add arrangements missing 4 colors (there are no arrangements missing 5 colors).


In short, from inclusion-exclusion principle, we have
$$X = \binom{50}{6}-\binom{5}{1}\binom{40}{6}+\binom{5}{2}\binom{30}{6}-\binom{5}{3}\binom{20}{6}+\binom{5}{4}\binom{10}{6}$$
There is, however, an easier way to solve this problem. If we select 6 balls so that there is at least one from each color, we'll have exactly one ball per color, except for one color which will have two.
So we have 5 ways to choose the color that has two balls, $10 \choose 2$ ways to choose those 2 balls, and then 10 ways per color to choose the balls from the remaining colors. Thus,
$$X = 5 \times {10 \choose 2} \times 10^4 = 5 \times 45 \times 10 000 = 2 250 000$$.
Thus, the sum we need to compute is 2 250 000.
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