Question 1174094: In how many 7 children can be arranged in a line, such that two particular children of them are never together?
a. 360
b. 480
c. 720
d. 1440
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Label the chidren as A,B,C,D,E,F,G
Let's say we want child A and child B to never be sitting together. That's the ultimate goal that we'll aim for.
However, for now we'll consider the opposite and count the number of ways that we can arrange the children such that A and B are always together.
What we'll do is remove both of these letters from the list, and replace them with H
So we go from
A,B,C,D,E,F,G
to
H,C,D,E,F,G
H is a placeholder for either A,B or B,A.
There are 6 letters in the set {H,C,D,E,F,G} so there are 6! = 6*5*4*3*2*1 = 720 ways to permute these items.
Since there are two ways to replace H, for any of the 720 permutations mentioned, this means there are overall 2*720 = 1440 different ways to have the children arranged such that child A and child B are always sitting next to each other.
However, we want them separated. So we must subtract that count from 7! = 7*6*5*4*3*2*1 = 5040 because this represents the total number of ways to arrange all seven kids without any restrictions.
Phrased another way:
answer we want = (number of ways to arrange all kids with no restrictions) - (number of ways to arrange the kids such that A,B are always together)
We could use algebra to tidy that equation up
x = number of ways to arrange all kids with no restrictions (A and B may or may not be sitting together)
y = number of ways to arrange the kids such that A,B are always together
z = number of ways to arrange the kids where A,B are never together
So based on that we can say,
z = x-y
Earlier we found that
x = 5040
y = 1440
So,
z = x-y
z = 5040-1440
z = 3600
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Answer: 3600
I think your teacher made a typo and meant to say 3600 for choice A, instead of 360.
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