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**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?