ScottTargetTestPrep wrote:
Bunuel wrote:
In how many ways can the crew of a ten oared boat be arranged, when of the ten persons available, 2 of whom can row only on the bow side and 3 of whom can row only on the stroke side?
A. \(\frac{10!}{2!*3!}\)
B. \(\frac{10!}{8!*7!}\)
C. \(\frac{5!}{3!*2!}\)
D. \(\frac{(5!)^3}{3!*2!}\)
E. \(\frac{5!}{8!*7!}\)
The two additional persons to row on the stroke side can be chosen in 5C2 = 5!/(3!*2!) ways. Notice that once this choice has been made, all the remaining crew must row on the bow side; therefore, there is only one way to choose the 5 people to row on the bow side.
The five people to row on the stroke side can be arranged in 5! ways. Similarly, the five people to row on the bow side can also be arranged in 5! ways. Therefore, the boat can be crewed in
5!/(3!*2!) * 5! * 5! = (5!)^3/(3!*2!)
ways.
Answer: D
ScottTargetTestPrep could you please explain where is my thinking flawed?
I select the two persons to row on the stroke side using K!/(N-K)! since I thought that different orders would have given me different combinations, so:
5!/3! * 3! * 5! since I also need to consider the possible combinations of the other three people on the stroke side and the 5 people on the bow side.
I actually calculated the value but it doesn't match
Could you explain please? Thanks