The starting line up for a basketball team of 5 players is to be selected from the team of 15 seniors and 10 juniors. The team is is set to pick a PG, SG, SF, PF, and C. How many ways are there to select the team if the PG has to be a senior and the SG has to be a junior?

So this is what I did.

  1. PG - Senior
  2. SG - Junior
  3. SF - 14 Seniors or 9 Juniors
  4. PF- 13 Seniors or 8 Juniors
  5. C - 12 Seniors or 7 Juniors

so $$14(13)(12)(9)(8)(7)= 1,100,736$$ possible ways to select the team

is this right?

  • $\begingroup$ It doesn't look like you quite got it. If the PG is a senior there are 15 options, and 10 options for the SG. Note that this leaves $9+14 = 23$ team members to fill in the other positions. $\endgroup$ – Mnifldz Mar 30 '16 at 18:02
  • $\begingroup$ So instead of looking at who is a senior and who is a junior, would you just do 23.22.21 =10,626 possible ways to select the team? $\endgroup$ – user323548 Mar 30 '16 at 18:08
  • $\begingroup$ No, you have to take into account the first two positions. Your comment shows the number of ways to fill the other three slots. $\endgroup$ – Mnifldz Mar 30 '16 at 18:09

First select the senior for the PG - 15 ways, then junior for the SG - 10 ways. and then you have $23\cdot22\cdot21$ ways to fill in the other positions.

In total this gives you:

$23\cdot22\cdot21\cdot15\cdot10 = 1593900$ ways to select the team


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