# Thread: Proof of k-multiset coefficients, part 1

1. ## Proof of k-multiset coefficients, part 1

This exercise outlines a bijective proof of the formula $\displaystyle \left(\left({n\atop k}\right)\right)= \left({k+n-1\atop k}\right)$ from section 1.1. Let A be the set of k-multisets taken from [n] and let B be the set of k-subsets of [k+n-1]. Assume that the k-multiset $\displaystyle \left\{ a_{1},a_{2},\ldots,a_{k}\right\}$ is written in nondecreasing order: $\displaystyle a_{1}\leq a_{2}\leq\ldots\leq a_{k}$. Define f: A $\displaystyle \longrightarrow$ B by
$\displaystyle f\left(\left\{ a_{1},a_{2},\ldots,a_{k}\right\} \right)=\left\{ a_{1},a_{2}+1,a_{3}+2,\ldots,a_{k}+n-1\right\}$ .

This function, and proof, is originally due to Euler.

Prove that the outputs of $\displaystyle f$ are indeed k-subsets of $\displaystyle [k+n-1]$ . This requires proof since it is not immediately clear from the definition of $\displaystyle f$.

Proof: Since $\displaystyle f(a_{i})=a_{i}+i-1, 1 \leq i \leq k$ and $\displaystyle a_{1}\leq a_{2}\leq\ldots\leq a_{k}$ then we know that the largest output of $\displaystyle f$ is $\displaystyle a_{k}+k-1$. Therefore $\displaystyle a_{1}\leq a_{i}+i-1\leq a_{k}+k-1, 1 \leq i \leq k$ for any output of $\displaystyle f$ . Since $\displaystyle a_{1}$ is the smallest member of [k+n-1] by definition, it follows that any $\displaystyle (a_{i}+i-1)\subseteq[k+n-1], 1 \leq i \leq k$ and that the outputs of $\displaystyle f$ are indeed k-subsets of [k+n-1].

2. Originally Posted by oldguynewstudent
This exercise outlines a bijective proof of the formula $\displaystyle \left(\left({n\atop k}\right)\right)= \left({k+n-1\atop k}\right)$ from section 1.1. Let A be the set of k-multisets taken from [n] and let B be the set of k-subsets of [k+n-1]. Assume that the k-multiset $\displaystyle \left\{ a_{1},a_{2},\ldots,a_{k}\right\}$ is written in nondecreasing order: $\displaystyle a_{1}\leq a_{2}\leq\ldots\leq a_{k}$. Define f: A $\displaystyle \longrightarrow$ B by
$\displaystyle f\left(\left\{ a_{1},a_{2},\ldots,a_{k}\right\} \right)=\left\{ a_{1},a_{2}+1,a_{3}+2,\ldots,a_{k}+n-1\right\}$ .

This function, and proof, is originally due to Euler.

Prove that the outputs of $\displaystyle f$ are indeed k-subsets of $\displaystyle [k+n-1]$ . This requires proof since it is not immediately clear from the definition of $\displaystyle f$.

Proof: Since $\displaystyle f(a_{i})=a_{i}+i-1, 1 \leq i \leq k$ and $\displaystyle a_{1}\leq a_{2}\leq\ldots\leq a_{k}$ then we know that the largest output of $\displaystyle f$ is $\displaystyle a_{k}+k-1$. Therefore $\displaystyle a_{1}\leq a_{i}+i-1\leq a_{k}+k-1, 1 \leq i \leq k$ for any output of $\displaystyle f$ . Since $\displaystyle a_{1}$ is the smallest member of [k+n-1] by definition, it follows that any $\displaystyle (a_{i}+i-1)\subseteq[k+n-1], 1 \leq i \leq k$ and that the outputs of $\displaystyle f$ are indeed k-subsets of [k+n-1].
b) Prove that f is a bijection.

Proof: Let $\displaystyle f(a_{i1})=f(a_{i2})$.We need to show that $\displaystyle a_{i1}=a_{i2}$. Now suppose that $\displaystyle a_{i1}\neq a_{i2}$, then we have two cases:

Case1: $\displaystyle a_{i1}<a_{i2}$. We know that $\displaystyle f(a_{i1})=a_{i1}+i1-1$ and $\displaystyle f(a_{i2})=a_{i2}+i2-1$. If $\displaystyle a_{i1}<a_{i2}$ then since the set is written in nondescending order, we know that $\displaystyle i1<i2$. If we add the inequalities together, we get $\displaystyle a_{i1}+i1<a_{i2}+i2$. Now subtract 1 from both sides preserving the inequality to obtain $\displaystyle a_{i1}+i1-1<a_{i2}+i2-1$ which is a contradiction.

Case2: $\displaystyle a_{i1}>a_{i2}$. Similarly, we have $\displaystyle i1>i2$, so we get $\displaystyle a_{i1}+i1>a_{i2}+i2$, and subtracting 1 from both sides we get $\displaystyle a_{i1}+i1-1>a_{i2}+i2-1$,which is again a contradiction.

Therefore, $\displaystyle a_{i1}=a_{i2}$which proves that $\displaystyle f$ is injective. Since $\displaystyle f$ is both injective and onto, it is a bijection. QED