Difference between revisions of "1958 AHSME Problems/Problem 49"
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==Solution 2 (Stars and Bars)== | ==Solution 2 (Stars and Bars)== | ||
− | Each term in the expansion of <math>(a+b+c)^{10}</math> will have the form <math>a^i \times b^j \times c^k</math>, where | + | Each term in the expansion of <math>(a+b+c)^{10}</math> will have the form <math>a^i \times b^j \times c^k</math>, where <math>0\le i, j, k\le 10</math> and <math>a+b+c=10</math>. So, we need to find the number of triplets of nonnegative integers <math>(a, b, c)</math> such that <math>a+b+c=10</math>. Using Stars and Bars, this value is <math>\binom{12}{2}=66</math>. |
==See also== | ==See also== |
Latest revision as of 00:59, 1 January 2024
Problem
In the expansion of there are
dissimilar terms. The number of dissimilar terms in the expansion of
is:
Solution
Expand the binomial with the binomial theorem. We have:
So for each iteration of the summation operator, we add k+1 dissimilar terms. Therefore our answer is:
Solution 2 (Stars and Bars)
Each term in the expansion of will have the form
, where
and
. So, we need to find the number of triplets of nonnegative integers
such that
. Using Stars and Bars, this value is
.
See also
1958 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 48 |
Followed by Problem 50 | |
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