# YADP (Yet Another Data Problem) :)

#### EddSjd

hello again! i believe i have gotten a liiiiittle better at data since this morning. i worked through all the questions in the chapter and came to the last one and got stuck.. again >.>

the question:

"a committee of three teachers are to select the winner from among ten students nominated for a special award. The teachers each make a list of their top three choices in order. The list have only one name in common, and that name has a different rank on each list. In how many ways could the teachers have made their lists?"

i cant get that answer.. >.> as far as ive done it, there are 3 different lists, with 6 spots left unfilled. So i tried 3*(9permutate6) but thats not even close...

#### skipjack

Forum Staff
The specification that "The lists have only one name in common" is ambiguous; the intended interpretation is that one name is common to all three lists, but no other name is repeated at all.

There are 10 possibilities for the common name, and 3 Ã— 2 ways in which the teachers can place it (since each teacher gives it a different place).

For each of those 10 Ã— 3 Ã— 2 cases, the first teacher's remaining places can be filled in 9 Ã— 8 ways, for each of which the second teacher's remaining places can be filled in 7 Ã— 6 ways, for each of which the third teacher's remaining places can be filled in 5 Ã— 4 ways.

Thus the required total is 10 Ã— 3 Ã— 2 Ã— 9 Ã— 8 Ã— 7 Ã— 6 Ã— 5 Ã— 4 = 10! = 3628800.

#### EddSjd

oh shoot i didnt even see the 10 possibilities...
but now i get it ty so much 8)