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Thread: Pascal

  1. #1
    Senior Member Sampras's Avatar
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    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 View Post
    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?
    Last edited by Krizalid; May 27th 2009 at 05:19 PM.
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  2. #2
    Senior Member
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    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.
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