Question 892694: Is this a trick question or does the 6 seats in the first row alter the answer?
In how many ways can 5 students from a class of 16 be arranged in the first row of seats. (There are 6 seats in the first row).
Help please.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website!
the number of ways you can get 5 students from 16 students is 16C5.
this does not take order into account.
this just gets you 5 students in each set without order.
then you want to arrange those 5 students in the 6 seats.
the number of possible sets of students is now 16C5 = 4368 possible sets of 5 that are each unique from each other.
each one of these sets needs to be arranged into the 6 seats.
take one at a time.
the vacant seat is equivalent to the 6th student that is added to each set.
therefore each set of 6 students can be arranged 6! times which is equal to 720 times.
the total number of possible arrangements will be 4368 * 720 = 3144960.
i tested this out with 2 out of 3 students being arranged in 3 seats.
it looks like the formula i gave you is correct.
here's the rationale.
the 3 students are abc (a is one of the students, b is one of the other students, c is the last of the 3 students).
the possible ways to get 2 out of the 3 students without order is 3C2 = 3.
those possible arrangements are:
ab
ac
bc
now each one of these sets of 2 students needs to be arranged in 3 seats.
that means one of the 3 seats will be vacant each time.
this effectively means that each set of 2 students will be arranged in 3! ways with the vacant seat representing the third student.
you have 3! ways for the first set of 2 students.
you have 3! ways for the second set of 2 students.
you have 3! ways for the third set of 2 students.
the total number of ways 2 students out of 3 can be arranged in 3 sets should be equal to 3C2 * 3! if this is correct.
that would be a total of 3*6 = 18 ways.
let's see if we can come up with those arrangements.
3 sets of students are ab, ac, bc.
we'll let v represent the vacant seat.
ab can be arranged in 3 seats in the following ways.
abv
avb
bav
bva
vab
vba
that's 6 possible arrangements for the first set of 2 students.
the second set of students is ac
they can be arranged in the following 6 ways.
acv
avc
cav
cva
vac
vca
the third set of students is bc
they can be arranged in the following 6 ways.
bcv
bvc
cbv
cvb
vbc
vcb
looks like the formula is correct and should be able to be applied to your larger problem.
the solution to your larger problems is 16C5 * 6! = 3144960.
i won't swear this is correct, but it appears to be correct if my assumptions about what is being asked are correct.
here's another way to look at it.
if it was only 5 seats in the first row, the solution would have been 16P5 because then you would want the 5 students with order.
16P5 = 524160
this is the same answer you would get if you took 16C5 and then multiplied it by 5!.
you would get 16C5 * 5! = 4368 * 120 = 524160.
you're doing the same thing in this problem except you are multiplying by 6! rather than 5!.
the combination formula and the permutation formula are related in the following way.
nCx = nPx / x!
nPx = nCx * x!
this problem appears to be an adaptation of that second formula with the additional wrinkle that instead of multiplying by x!, you will be multiplying by (x+1)! that was required because of that extra seat that would be vacant each time.
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