10 kids are going to stand in a row. They can stand in any

Question

10 kids are going to stand in a row. They can stand in any order except for the following rule: James should stand left to Andrew, Andrew should stand left to Mark. How many different variants for kids to take places in a row are possible?

(A) 10!/3!7!
(B) 10!/7!
(C) 3*7!
(D) 10!/3!
(E) 3!*7!

jingy77 · Answer

Anyone? I would like to see how this is done.

walker · Answer

D

The total number of options =  P^{10}_{10}

for James,Andrew, and Mark we have  P^3_3  possible mutual placements but only one is satisfies the condition.

Therefore, N= \frac{P^{10}_{10}}{P^3_3}=\frac{10!}{3!}

bmwhype2 · Answer

it says J A M in that order...
i dont understand why we are dividing by 3...

Formermember99 · Answer

If we assume that JAM form a group -- imagine them holding hands -- then we could think of them as a single unit, and the other 7 children can be arranged around them, so it would really just be 8!.

However, since that's not even an option, I guess the question is meant to say, there can be any number of children in between them... so, J... A... M. That's trickier.

Let's start with an easier version. What if there were 10 children, and we only had to worry about J being to the left of A? Then, for every arrangement of the children where J was to the left of A, there would a mirror image version where J was to the right of A, right? That means that half the options wouldn't fit the criteria. So of the original 10! options, only half could be kept. 10!/2 would be the answer.

But... what if there are three? Well, now we can't quite think about mirror images. Now we have to think about it further. When there was only J and A to worry about, we divided out half the possibilities because there are two ways to arrange J and A for every arrangement of the other 8 kids, and only one of those is allowed. So, divide by 2. Ok, so when there are J and A and M to worry about, we have to divide out more of the possibilities. We have to get rid of 1/6th of the possibilities, because for every arrangement of the other kids, J and A and M could be arranged in 3x2x1 ways.... and only one of those is allowed. So, we divide by 6, to keep one out of every six.

So, long and short, 10!/3!. But I hope that's clearer as to why.

sreehari · Answer

OA is D. Although I guess the answer, didn't understand the rational behind the denominator, 3!.

walker, Lorettagrace - Thanks for the explanations.

Formermember99 · Answer

It's not a rule. It's responding to the question. The question explicitly states that only J... A... M is an acceptable order. So for every arrangement of the other children, the only acceptable one is the one that has J.. A... M in that order. Only one in six (the six comes from 3! and the the 3 in 3! comes from the fact that there are three people in the group J...A... M) will have that arrangement.

If you don't see why ...

Let's call the other seven children 1 2 3 4 5 6 7.

Here's one random arrangement of them, with J and A and M in it.

1 J 2 3 A 4 5 6 M 7

Here's some similar arrangements:

1 J 2 3 A 4 5 6 M 7
1 J 2 3 M 4 5 6 A 7
1 A 2 3 J 4 5 6 M 7
1 A 2 3 M 4 5 6 J 7
1 M 2 3 J 4 5 6 A 7
1 M 2 3 A 4 5 6 J 7

What's the same is the other 7 children. What's different is the J... A... M.

If you "tune out" the other children, you see what's happening. This is all 6 permuations of JAM.

JAM
JMA
MAJ
MJA
AJM
AMJ

Of those six arrangements (and you can find that by 3! if you don't want to list it), only one is acceptable.

The same will be true of any arrangement of the other 7 children.

So, there are a grand total of 10! ways to arrange all 10 children. If we divide by 3!, we'll get rid of the unacceptable arrangements.

I believe my original explanation is fairly clear if you begin at the beginning. If you understand why you'd divide by two if we're only talking about J and A, then you can apply that understanding to why you'd divide by 6, aka, 3!, when we're talking about J and A and M.

I can't explain it any more without repeating myself, so I hope that's enough. If it isn't, maybe you're one of the people who likes to just apply formulas without understanding them, and that's ok, a lot of people are like that.

bmwhype2 · Answer

thanks for writing it out. if we divide by the number of unfavorable outcomes, why is it 3! and not 5 since 1 of 6 is acceptable?

Formermember99 · Answer

Since 1 in 6 is acceptable, if we divide by 6, what's left is acceptable. 3! is the same thing as 6.

10 kids are going to stand in a row. They can stand in any

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