Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1958 AHSME Problems/Problem 49"

(→‎Problem)
(→‎Solution 2 (Stars and Bars))
 
(10 intermediate revisions by the same user not shown)
Line 19: Line 19:
 
<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 \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 23:59, 31 December 2023

Problem

In the expansion of $(a + b)^n$ there are $n + 1$ dissimilar terms. The number of dissimilar terms in the expansion of $(a + b + c)^{10}$ is:

$\textbf{(A)}\ 11\qquad \textbf{(B)}\ 33\qquad \textbf{(C)}\ 55\qquad \textbf{(D)}\ 66\qquad \textbf{(E)}\ 132$

Solution

Expand the binomial $((a+b)+c)^n$ with the binomial theorem. We have:

\[\sum\limits_{k=0}^{10} \binom{10}{k} (a+b)^k c^{10-k}\]

So for each iteration of the summation operator, we add k+1 dissimilar terms. Therefore our answer is:

\[\sum\limits_{k=0}^{10} k+1 = \frac{11(1+11)}{2} = 66 \to \boxed{\textbf{D}}\]

Solution 2 (Stars and Bars)

Each term in the expansion of $(a+b+c)^{10}$ will have the form $a^i \times b^j \times c^k$, where $0\le i, j, k\le 10$ and $a+b+c=10$. So, we need to find the number of triplets of nonnegative integers $(a, b, c)$ such that $a+b+c=10$. Using Stars and Bars, this value is $\binom{12}{2}=66$.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1958_AHSME_Problems/Problem_49&oldid=209315"