Difference between revisions of "1958 AHSME Problems/Problem 49"
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<cmath>\sum\limits_{k=0}^{10} k+1 = \frac{11(1+11)}{2} = 66 \to \boxed{\textbf{D}}</cmath> | <cmath>\sum\limits_{k=0}^{10} k+1 = \frac{11(1+11)}{2} = 66 \to \boxed{\textbf{D}}</cmath> | ||
| + | ==Solution 2 (Stars and Bars)== | ||
| + | |||
| + | Each term in the expansion of <math>(a+b+c)^10</math> will have the form <math>a^i\timesb^jtimesc^k</math>, where $0\lei, j, k/le10 | ||
==See also== | ==See also== | ||
Revision as of 23:46, 31 December 2023
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:
![\[\sum\limits_{k=0}^{10} \binom{10}{k} (a+b)^k c^{10-k}\]](http://latex.artofproblemsolving.com/e/7/2/e725bca22403df696716c07255dd306e9dcd4333.png)
So for each iteration of the summation operator, we add k+1 dissimilar terms. Therefore our answer is:
![\[\sum\limits_{k=0}^{10} k+1 = \frac{11(1+11)}{2} = 66 \to \boxed{\textbf{D}}\]](http://latex.artofproblemsolving.com/b/d/c/bdc92d46f69d4d77597bf6b65ef54c6d80639b20.png)
Solution 2 (Stars and Bars)
Each term in the expansion of 
will have the form $a^i\timesb^jtimesc^k$ (Error compiling LaTeX. Unknown error_msg), where $0\lei, j, k/le10
See also
| 1958 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 48 |
Followed by Problem 50 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 • 41 • 42 • 43 • 44 • 45 • 46 • 47 • 48 • 49 • 50 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
