Difference between revisions of "University of South Carolina High School Math Contest/1993 Exam/Problem 14"
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For the nine slots, we have to choose two spots for each of our pairs <math>(1,2), (3,4), (5,6)</math>. There are <math>{9\choose 2}{7\choose 2}{5\choose 2}</math> ways to do this, but we must multiply this by <math>3!</math> because there are three different pairs that we can choose in different "orders" for the <math>{9\choose 2}, {7\choose 2}, {5\choose 2}</math>. We also need to multiply by <math>3!</math> again for the remaining three slots for <math>7,8,9</math>. Thus the total number of ways is <math>3!{9\choose 2}{7\choose 2}{5\choose 2}3!</math>, which after simplifying is choice <math>(a)</math>. | For the nine slots, we have to choose two spots for each of our pairs <math>(1,2), (3,4), (5,6)</math>. There are <math>{9\choose 2}{7\choose 2}{5\choose 2}</math> ways to do this, but we must multiply this by <math>3!</math> because there are three different pairs that we can choose in different "orders" for the <math>{9\choose 2}, {7\choose 2}, {5\choose 2}</math>. We also need to multiply by <math>3!</math> again for the remaining three slots for <math>7,8,9</math>. Thus the total number of ways is <math>3!{9\choose 2}{7\choose 2}{5\choose 2}3!</math>, which after simplifying is choice <math>(a)</math>. | ||
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[[Category:Intermediate Combinatorics Problems]] | [[Category:Intermediate Combinatorics Problems]] | ||
Revision as of 12:18, 23 July 2006
Problem
How many permutations of 1, 2, 3, 4, 5, 6, 7, 8, 9 have:
- 1 appearing somewhere to the left of 2,
- 3 somewhere to the left of 4, and
- 5 somewhere to the left of 6?
For example, 8 1 5 7 2 3 9 4 6 would be such a permutation.
Solution
For the nine slots, we have to choose two spots for each of our pairs 
. There are 
ways to do this, but we must multiply this by 
because there are three different pairs that we can choose in different "orders" for the 
. We also need to multiply by again for the remaining three slots for 
. Thus the total number of ways is 
, which after simplifying is choice 
.
