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I believe I have this figured out but I'm no math wiz and I need someone to check that my work is correct.

I have a fantasy football lineup consisting of 6 positions with a total of 9 players used in each lineup. Each position uses one player at a time except for Running Back and Wide Receiver.

  • Quarterback (QB) $\times 1$
  • Running Back (RB) $\times 2$
  • Wide Reciever (WR) $\times 3$
  • Tight End (TE) $\times 1$
  • Flex (FLEX) $\times 1$
  • Defense (DST) $\times 1$

The Flex position will be chosen from one of the players left over in the RB, WR or TE positions.

Lets say I have a player pool consisting of 5 of each of the positions except for the FLEX spot since that player will come from the RB, WR, TE pool.

The first thing we need to do is calculate the total number of combinations of 2 RB and 3 WR from each of their pools.

$n$ choose $r$ = $^{n}C_r$

$^{RB}C_r =$ $ ^{5}C_2 = 10$ RB Combinations (RBC)

$^{WR}C_r =$ $ ^{5}C_3 = 10$ WR Combinations (WRC)

Now we need to get the number of players left over from the possible FLEX positions so we add those positions together but then subtract 6 for the 6 players we will have already chosen to use in that lineup.

$((RB+WR+TE)-6) = ((5+5+5)-6) = 9$ FLEX Choices.

Now we have the numbers we need to do the final calculation. We simply multiply to find the total number of different lineup combinations.

$(QB*RBC*WRC*TE*FLEX*DST) = (5*10*10*5*9*5) = 112,500$

Can someone please tell me if this is correct or if I have missed something?

I'm sorry if I have messed any of the math notation up as I said I'm not a wiz. I have a high school education and my algebra teacher was an alcoholic who actually tried to fail me because he couldn't average my grade correctly.

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This is fine if the flex player is labeled FLEX and cannot swap with one of the players from his position. To understand the other case suppose you had tight ends named A,B,C,D, and E. Is a team with tight end A and flex C different from a team with tight end C and flex A? If you just list the players on your team you get the same list either way.

If the player is not labeled, you have double counted all the teams where the flex player is a tight end, counted four times the teams where the flex player is a wide receiver, and counted three times the teams where the flex player is a running back. You will have to separate your calculation into cases.

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  • $\begingroup$ I had not considered that. How would you go about separating them into cases? $\endgroup$ – Shre Nov 11 '15 at 21:41
  • $\begingroup$ If the flex is a tight end you select two tight ends out of five, so there are $5*10*10*10*5=25000$ teams. If the flex is a wide receiver you select four wide receivers out of 5, so there are $5*10*5*5*5=6250$ teams. For flex a running back you choose three out of five, so there are $5*10*5*5=12500$ teams. Total is $43750$ teams. $\endgroup$ – Ross Millikan Nov 11 '15 at 21:46
  • $\begingroup$ Wouldn't running back flex be $5*10*5*5*5 = 6250$ for a total of 37500 teams? $\endgroup$ – Shre Nov 11 '15 at 21:57
  • $\begingroup$ No, it would be $5*10*10*5*5$. You choose three running backs and three wide receivers. I missed typing one of the $10$'s $\endgroup$ – Ross Millikan Nov 11 '15 at 21:59

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