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n2m or m2n ? (Posted on 2014-07-16) Difficulty: 2 of 5
Let P1 be a polynomial of n terms raised to the mth power.
Let P2 be a polynomial of m terms raised to the nth power.

When expanded, which of the two will yield more terms?
Please explain your choice.

No Solution Yet Submitted by Ady TZIDON    
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Solution re(2): Trivial solution? | Comment 5 of 6 |
(In reply to re: Trivial solution? by Ady TZIDON)

So if we have (a + b + c)^n, then there will be all of the following terms in the expansion:

a^n

b^n

c^n

a^(n-1)*b

a^(n-1)*c

a^(n-2)*b*c

a^(n-2)*b^2

etc.

Basically, every combination of a, b, and c such that their exponents add up to n.  There are (n+2) choose 2 such combinations, since we have to order the n powers and 2 partitions between the three components of the polynomial. 

So if a polynomial of m terms is raised to the nth power, the number of terms in the expansion will be (n+m-1) choose (m-1).

So we have the number of terms in the expansion of P1, N(P1) = (n+m-1) choose (n-1) = (n+m-1)! / m!(n-1)!

and N(P2) = (n+m-1) choose (m-1) = (n+m-1)! / n!(m-1)!

This simplifies nicely to N(P1) = N(P2) * n / m

So if 2 < n < m, then n/m < 1 and thus N(P1) < N(P2) for all n,m (which agrees with my "trivial" observation).

 

Edited on July 16, 2014, 3:46 pm
  Posted by tomarken on 2014-07-16 14:05:56

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