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Sample Question 10:
Step 1: Notice that the numbers from 8 through 14 must each form distinct pairs because they are all larger than any number from 1 through 7. Also, number 7 must pair with 14.
Step 2: Number 6 can pair with either 12 or 13. We can consider these possibilities as two separate cases:
Case 1: If 6 pairs with 12, then 5 can pair with one of 10,11,13. Once these pairs are formed, the remaining numbers 1,2,3,4 can be freely paired with any of the remaining higher numbers. This case yields \(3*4! = 72\) valid ways to form the pairs.
Case 2: If 6 pairs with 13, then 5 can pair with one of 10,11,12. Again, the remaining numbers 1,2,3,4 can be freely paired with any remaining higher numbers. This case also yields \(3*4!=72\) valid ways.
Step 3: Adding together the results from the two cases, we find a total of \(72+72=144\) ways to form the pairs such that the greater number is at least twice the lesser number.
Therefore, the answer is \(\boxed{\textbf{(E) } 144}\).
How many ways are there to split the integers \(1\) through \(14\) into \(7\) pairs such that in each pair, the greater number is at least \(2\) times the lesser number?
\(\textbf{(A) } 108 \qquad \textbf{(B) } 120 \qquad \textbf{(C) } 126 \qquad \textbf{(D) } 132 \qquad \textbf{(E) } 144\)
Answer Keys
Question 10: ESolutions
Question 10Step 1: Notice that the numbers from 8 through 14 must each form distinct pairs because they are all larger than any number from 1 through 7. Also, number 7 must pair with 14.
Step 2: Number 6 can pair with either 12 or 13. We can consider these possibilities as two separate cases:
Case 1: If 6 pairs with 12, then 5 can pair with one of 10,11,13. Once these pairs are formed, the remaining numbers 1,2,3,4 can be freely paired with any of the remaining higher numbers. This case yields \(3*4! = 72\) valid ways to form the pairs.
Case 2: If 6 pairs with 13, then 5 can pair with one of 10,11,12. Again, the remaining numbers 1,2,3,4 can be freely paired with any remaining higher numbers. This case also yields \(3*4!=72\) valid ways.
Step 3: Adding together the results from the two cases, we find a total of \(72+72=144\) ways to form the pairs such that the greater number is at least twice the lesser number.
Therefore, the answer is \(\boxed{\textbf{(E) } 144}\).