Using combinatorial methods, how can I solve the following problems?

  1. What value of $m$ maximizes: $$\sum_{i=0}^{n}\binom{10}{i}.\binom{20}{m-i}?$$

  2. What is the value of the sum: $$\binom{30}{0} \cdot \binom{30}{10}-\binom{30}{1} \cdot \binom{30}{11}+ \cdots +\binom{30}{20} \cdot \binom{30}{30}?$$

I have solved these questions using the expansion of $\bf{(1+x)^n}$ and $\bf{(1-x)^n}$.

But someone told me that we can easily solve them in a smaller number of steps using combinatorial methods. Can anyone explain this?


For (1): How many ways can you pick $m$ objects from 30 objects? You can pick $i$ from the first $10$ objects and $m-i$ from the next 20...

For the second one:


Think about it this way: You have 30 blue objects and 30 red objects.

Then the sum represents the number of way of choosing 20 objects so that the number of red and blue is even - the number of way of choosing 20 objects so that the number of red and blue is odd.

You should be able for each of the two parts to figure the right formula and get an easy closed form.

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  • $\begingroup$ Thanks N.S, Means Taking $m$ objects from Total $30$ objects $\displaystyle = \binom{30}{m}$. which is equal to taking $i$ from $10$ and Then $m-i$ from Next $20$ which is $\displaystyle \binom{10}{i}.\binom{20}{m-i}$ $\endgroup$ – juantheron Jan 22 '13 at 3:28
  • $\begingroup$ for (2) one: selecting $2i$ blue balls from $30$ blue balls $\displaystyle = c\binom{30}{2i}$ and selcecting $20-2i$ red balls from $30$ red balls $ \displaystyle = \binom{30}{20-2i}$ Which is equal to Selecting $20$ balls from Total $30$ balls So $\displaystyle \sum_{i=0}^{10}\binom{30}{2i}.\binom{30}{20-2i} = \binom{60}{20}$ Is is Right or Not If not then would you like to explain it to me. Thanks $\endgroup$ – juantheron Jan 22 '13 at 3:54
  • $\begingroup$ For (1) I have Got it $\displaystyle \binom{30}{m} =$ Maximum, When$m=15$\\\\ for (2) Ans given is $\displaystyle \binom{30}{10}$ but I am Getting $\displaystyle \binom{60}{10}$ . So where i have done Mistake, Thanks $\endgroup$ – juantheron Jan 22 '13 at 17:32
  • $\begingroup$ for the second the mistake you did is that the fomula is not sellecteing 20 from 60. It is selecting 20 from 60, with even red and even blue. Which means selecting 10 pairs from 30 pairs ;) $\endgroup$ – N. S. Jan 22 '13 at 22:31

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