Factory X and factory Y are among the 5 factories in a certain manufac

One way to think about the problem:

Without the restriction there would be 5! = 120 ways the factories could be sequenced (you could either use the permutations formula, or use slots: 5 options for the first slot, then, once the first slot is filled, 4 options for the second, then, once the first two slots are filled, three options for the third and so own).

So without the restriction there would be 120 ways.

In half of those 120 ways X will be in front of Y and in the other half Y will be in front of X. (You might think something like: well neither of them is any more likely to be earlier than the other, so half the time X will be earlier and half the time Y will be earlier)

120/2 = 60

So E.

Hi Amberbajaj02,

This is similar to a question, where it says, There are five persons (A, B, C, D and E) standing in a row, in how many different arrangements person A will be ahead of person B.

As somebody replied above, it’s a symmetrical case, where in half of the total arrangements A would be ahead of B.

So here total arrangements of 5 things, 5! = 120. So In half of the cases he would visit “X” ahead of Y. so that is 60.

So another way of looking at this question is,

There are five slots,

B _ _ _ _ : So here A could be any of the four places, and the remaining three persons can be arranged in 3!. So that is 4 * 3!.

or

_ B _ _ _ : So here A could be any of the three places, and the remaining three persons can be arranged in 3!. So that is 3 * 3!.

or

_ _ B _ _ : So here A could be any of the two places, and the remaining three persons can be arranged in 3!. So that is 2 * 3!.

or

_ _ _ B _ : So here A could be only one place, and the remaining three persons can be arranged in 3!. So that is 1 * 3!.

So, total (4+3+2+1)*3!

10* 3! = 10 * 6 = 60.

So the answer is E.

Hope this is clear.

PS: I still feel that question should have framed better.
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There are 10 possible arrangements in which X is ahead of Y.

Then I calculated the possible arrangements of the remaining numbers, which is 3 * 2 * 1 = 6

6 * 10 = 60

Is this correct?

Bunuel

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Yes, however, the easiest method is just to do 5!/2. The 5! part counts all the ways to arrange 5 factories. When we divide by 2, it's like splitting these ways into two groups: in one group, X is ahead of Y, and in the other half, it's the opposite.

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Hope it helps.
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