Question 924409: A group of 12 people need to form a line. The line will consist of exactly 9 of the people. Person X and Person Y have to be either third or fourth in line. How many different orders are possible?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! what i think is:
if x and y have to occupy the third and fourth slot, that leaves 7 slots for the other 10 people.
instead of 12P9, you get 10P7 * 2P2
10P7 = 10*9*8*7*6*5*4 = 603800
2P2 = 2*1 = 2
10P7 * 2P2 = 603800 * 2 = 1209600
the 10P7 takes care of the number of arrangements for the 10 remaining people in 7 remaining slots.
the 2P1 takes care of the number of arrangements for the 2 people in 2 slots.
to see this with smaller numbers, consider 6 people for 4 slots where x and y have to be in position 1 and 2.
instead of 6P4 = 360 possible arrangements, you have 2P2 * 4P2 = 24 possible arrangements.
let the people be xy1234
x is a person, y is a person, 1 is a person, 2 is a person, etc.
x and y have to be in positions 1 and 2, so those positions are filled.
the possible positions are:
xy12
xy13
xy14
xy23
xy24
xy34
yx12
yx13
yx14
yx23
yx24
yx34
xy21
xy31
xy41
xy32
xy42
xy43
yx21
yx31
yx41
yx32
yx42
yx43
that's a total of 24.
the formula appears to work so it can be applied to the larger problem.
the solution appears to be:
10P7 * 2P2 = 1209600
x and y have to be in positions 2 and 3
that leaves only 1 position for a and b and c
the possible arrangements are:
axy
bxy
cxy
ayx
byx
cyx
there are no other possible arrangements.
|
|
|
