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I know how to do the problem conceptually, and proving the equality by computing isn't hard, but we are required to use a bijection to prove this, any ideas please?

https://i.stack.imgur.com/Q60pM.jpg

RobPratt
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Sam
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  • To choose $k$ things from $m+n$, you can choose $i$ things from the first $m$ and then $k-i$ things from the last $n$. – ajotatxe Oct 14 '19 at 13:48
  • yes, i realize that, we assume we have m balls and n balls and we want to pick k of them, we pick i from m, then we have to pick k-i from n, but how to write a bijection for this is my issue – Sam Oct 14 '19 at 13:50
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    I will point out that once you and/or your audience are more familiar with this notion of bijective proof, it is common to leave out the rigorous details of what the explicit bijection is and the proof of why it is a bijection as it is the "obvious" choice in each case, the only difficulty being having to come up with the exact sequence of symbols to get across the meaning that you want. – JMoravitz Oct 14 '19 at 14:02

2 Answers2

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HINT:

Try to translate the formulas into combinatorical concepts i.e.:

$$ \binom{a}{b} $$

means the number of ways to choose $b$ things out of $a$ things without caring about the order of the choices.

Vinyl_cape_jawa
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  • i understand that idea, we need to pick k items from m and n, we pick i (such that i is a natural number smaller than k) from m, meaning we pick k-i from n. but my issue is, how does the concept of bijection come into play here – Sam Oct 14 '19 at 13:52
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Let $U$ be the set $\{1,2,\ldots,m+n\}$, $V=\{1,\ldots,m\}$ and $W=\{m+1,\ldots,m+n\}$.

Now define $\mathcal P_j(A)$ as the set of subsets of some set $A$ that have exactly $j$ elements.

Consider $$f:\bigcup_{i=0}^k[\mathcal P_i(V)\times\mathcal P_{k-i}(W)]\to\mathcal P_k(U)$$ as $$f(A,B)=A\cup B$$

ajotatxe
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