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Math Help - Pascal

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

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

    (ii) Similarly,  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  53 cards?

    Its just  \binom{53}{5} ?

    3. Let  r_i be a positive integer,  1 \leq i \leq k . If  n = r_1+r_2+ \cdots + r_k , prove that  \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  0,0,1,3,6,10,15, \ldots from the third column of Pascal's triangle. Starting with  n=1 , the n \text{th} term of the sequence is  a_n = C(n,2) . Prove that for all  n \geq 0 , (i)  a_{n+1}-a_n = n and (ii)  a_{n+1}+a_n = n^2 .

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

    (ii) Similarly,  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  53 cards?

    Its just  \binom{53}{5} ?

    3. Let  r_i be a positive integer,  1 \leq i \leq k . If  n = r_1+r_2+ \cdots + r_k , prove that  \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:  \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  r_1 subsets, the number of  r_2 subsets, etc....?

    Because  C(n,r) is interpreted as how many ways we can choose an  r -subset from  n elements. In the same way, can we use this interpretation for the multinomial coefficient?
    Last edited by Krizalid; May 27th 2009 at 06:19 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 a_n=\frac12n(n+1). So 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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