six elves are lined up along a workbench what is the probablitiy that the shortest elf is next to the tallest?

Question

also, what is the probablity that the smallest is on the left and the tallest is on the right?

Answer 1

5!(2!)/6! = 1/3



Second question: 4!/6! = 1/30

Answer 2

For the first part, picture the tallest elf (who *must* be somewhere in the line!). Assume this elf has equal probability of being at any one of the six places in the line. For 4 of those six places (2/3 of the time), the tallest elf has an elf on each side, so 2 of the remaining 5 elves are next to the tallest. So the probability that the shortest elf is next to the tallest for those four places is 2/3 * 2/5 = 4/15. For the other two places (1/3 of the time), there is only one elf next to the tallest, so for those two places, the probability of the shortest being next to the tallest is 1/3 * 1/5 = 1/15. Adding the two probabilities we get the total probability of the shortest next to the tallest as P1 = 4/15 + 1/15 = 5/15 = 1/3.



You could also do this by noting that there are 10 different ways that the tallest and shortest elves can stand adjacent:

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And for each of those 10 ways, each of the four other elves can stand in 4! different setups, so there ar 10 * 4! arrangements that match the question, out of 6! total ways the elves could be standing. This means P1 = (10 * 4!) / 6! = 10 / (5 * 6) = 1/3.



For the second part, we said above that there are 6! permutations for the elves' positions, and if we put the smallest on the left and the tallest on the right, there are 4! permutations for where the other elves can stand. So the probability that the smallest is on the left and the tallest is on the right is P2 = 4! / 6! = 1 / (5 * 6) = 1/30.

Answer 3

Imagine six positions.



The shortest elf is in position 1. The tallest elf is next to him, in position 2. With this orientation there are 4x3x2x1 (4!)=24 different ways the last four elves can be lined up. This is because four different elves can be in the first position, three in the second, two in the third, and one in the last. It might take a while to get, that's alright. Anyway, this means there are 24 different ways that the elves can be lined up with the shortest and tallest beside each other IN THAT ORDER AT THAT END. Remember that.



Now imagine the two elves moving together down the line of elves, switching positions, and then moving back. This will give us all cases where the two elves are together. There are 10 cases.



If we do the same calculation for each one, there are still 24 possibilities for how the other elves are arranged for each position that they are in.



This means that there are 10x24=240 different cases where the shortest is next to the tallest.



Now the probability needs to be calculated. To do this we divide the number of cases of interest (240) by the total possible number of arrangements, whether the elves are together or not. To do this take 6!, with the same reasoning as taking 4! before. This should give a total number of 720 permutations (or cases).



Now, 240/720 = what? I'll let you figure the answer out :)

Answer 4

If the shortest elf is at one end or the other (2/6 of the possible arrangements) the chances are 1/5 that the tallest elf is next to him.



If the shortest elf is in positions 2-5 (4/6 of the possible arrangements) then the tallest elf has a 2/5 chance of being next to him.



So (2/6)*(1/5) + (4/6)*(2/5)



Lots of ways to do the 2nd one. For instance, there are 6! possible arrangements of all 6, but if you fix those two elves, then there are only 4! ways to rearrange the remaining elves. So 4!/6! arrangements meet the conditions.

Answer 5

1 in 5 1 in 10

Answer 6

1/6