# Pascal

• May 27th 2009, 03:27 PM
Sampras
1. Consider the sequence $\displaystyle 0,0,1,3,6,10,15, \ldots$ from the third column of Pascal's triangle. Starting with $\displaystyle n=1$, the $\displaystyle n \text{th}$ term of the sequence is $\displaystyle a_n = C(n,2)$. Prove that for all $\displaystyle n \geq 0$, (i) $\displaystyle a_{n+1}-a_n = n$ and (ii) $\displaystyle a_{n+1}+a_n = n^2$.

(i) So $\displaystyle a_{n+1}-a_n = C(n+1,2) - C(n,2)$. And $\displaystyle C(n+1,2) = C(n,1) + C(n,2)$. Thus $\displaystyle a_{n+1}-a_n= C(n,1) = n$.

(ii) Similarly, $\displaystyle a_{n+1}+a_n = C(n,1)+2C(n,2) = n^2$.

Are these correct?

2. Poker is sometimes played with a joker. How many different five-card poker hands can be "chosen" form a deck of $\displaystyle 53$ cards?

Its just $\displaystyle \binom{53}{5}$?

3. Let $\displaystyle r_i$ be a positive integer, $\displaystyle 1 \leq i \leq k$. If $\displaystyle n = r_1+r_2+ \cdots + r_k$, prove that $\displaystyle \binom{n}{r_1,r_2, \ldots, r_k} = \binom{n-1}{r_1-1,r_2, \ldots, r_k} + \binom{n-1}{r_1,r_2-1, \ldots, r_k} + \cdots + \binom{n-1}{r_1,r_2, \ldots, r_k-1}$.

So we can use both algebraic and combinatorial arguments for this?

Quote:

Originally Posted by Sampras
1. Consider the sequence $\displaystyle 0,0,1,3,6,10,15, \ldots$ from the third column of Pascal's triangle. Starting with $\displaystyle n=1$, the $\displaystyle n \text{th}$ term of the sequence is $\displaystyle a_n = C(n,2)$. Prove that for all $\displaystyle n \geq 0$, (i) $\displaystyle a_{n+1}-a_n = n$ and (ii) $\displaystyle a_{n+1}+a_n = n^2$.

(i) So $\displaystyle a_{n+1}-a_n = C(n+1,2) - C(n,2)$. And $\displaystyle C(n+1,2) = C(n,1) + C(n,2)$. Thus $\displaystyle a_{n+1}-a_n= C(n,1) = n$.

(ii) Similarly, $\displaystyle a_{n+1}+a_n = C(n,1)+2C(n,2) = n^2$.

Are these correct?

2. Poker is sometimes played with a joker. How many different five-card poker hands can be "chosen" form a deck of $\displaystyle 53$ cards?

Its just $\displaystyle \binom{53}{5}$?

3. Let $\displaystyle r_i$ be a positive integer, $\displaystyle 1 \leq i \leq k$. If $\displaystyle n = r_1+r_2+ \cdots + r_k$, prove that $\displaystyle \binom{n}{r_1,r_2, \ldots, r_k} = \binom{n-1}{r_1-1,r_2, \ldots, r_k} + \binom{n-1}{r_1,r_2-1, \ldots, r_k} + \cdots + \binom{n-1}{r_1,r_2, \ldots, r_k-1}$.

So we can use both algebraic and combinatorial arguments for this?

For (3) would you just use the definition of multinomial coefficient: $\displaystyle \binom{n}{k_{1}, k_{2}, \ldots, k_m} = \frac{n!}{k_{1}!k_{2}! \cdots k_{m}!}$ for an algebraic argument?

And for a combinatorial argument, you would consider the number of $\displaystyle r_1$ subsets, the number of $\displaystyle r_2$ subsets, etc....?

Because $\displaystyle C(n,r)$ is interpreted as how many ways we can choose an $\displaystyle r$-subset from $\displaystyle n$ elements. In the same way, can we use this interpretation for the multinomial coefficient?
• May 28th 2009, 05:19 PM
Media_Man
Famous Sequence
1. The sequence you are referring to is given by $\displaystyle a_n=\frac12n(n+1)$. So $\displaystyle a_{n-1}+a_n=\frac12(n(n+1)+n(n+1))=\frac12(n^2+n+n^2-n)=n^2$ . I'll let you work the other one out.

2. Yes.