Difference between revisions of "2007 AMC 10A Problems/Problem 12"
(→Solution) |
Pi is 3.14 (talk | contribs) (→Solution) |
||
| Line 3: | Line 3: | ||
<math>\text{(A)}\ 56 \qquad \text{(B)}\ 58 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 62 \qquad \text{(E)}\ 64</math> | <math>\text{(A)}\ 56 \qquad \text{(B)}\ 58 \qquad \text{(C)}\ 60 \qquad \text{(D)}\ 62 \qquad \text{(E)}\ 64</math> | ||
| + | |||
| + | == Video Solution == | ||
| + | https://youtu.be/0W3VmFp55cM?t=3352 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
== Solution == | == Solution == | ||
Revision as of 22:26, 16 January 2021
Problem
Two tour guides are leading six tourists. The guides decide to split up. Each tourist must choose one of the guides, but with the stipulation that each guide must take at least one tourist. How many different groupings of guides and tourists are possible?
Video Solution
https://youtu.be/0W3VmFp55cM?t=3352
~ pi_is_3.14
Solution
Each tourist has to pick in between the 
guides, so for 
tourists there are 
possible groupings. However, since each guide must take at least one tourist, we subtract the cases where a guide has no tourist. Thus the answer is 
.
Solution 2
Without loss of generality, let's call one of the tour guides tour guide A, and the other tour guide B. To count the number of total groupings of guides and tourists possible, we can count the number of ways some number of tourists go to tour guide A. Thus, we can see that the total groupings is:
![\[\binom{6}{1} + \binom{6}{2} + \binom{6}{3} + \binom{6}{4} + \binom{6}{5} = 62\]](http://latex.artofproblemsolving.com/3/c/a/3cabf2e89a63ed0a183a744842d50e273f53c088.png)
See also
| 2007 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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 | ||
| All AMC 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
