# 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

• No money for a calculator ? Use wolframalpha maybe? – Henno Brandsma Dec 22 '18 at 10:27
• I bealive D Tiwari want some nice combinatorial simplification of this exspresion. Don't understand why is this voted for close. – Aqua Dec 22 '18 at 13:46
• Yes greedoid i want combinational argument. – DXT Dec 22 '18 at 14:26
• Anyway it's a finite sum. Maybe a related to it but somehow a general one will be interesting. – Felix Marin Dec 22 '18 at 16:36
• Answer - 2250000 – ablmf Dec 24 '18 at 15:19

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.

1. We count how many ways we can choose 6 balls. That is $$50 \choose 6$$
2. 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.
3. 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.
4. 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.