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If we have 4 items and 11 spots, then if we randomly and independently assign 1 spot to each item, how many ways the 4 items can be arranged in 11 spots. There is a possibility that you may arrange different item in the same spot also.

Here is my try on this:

If we arrange no two items in the same spot, then no. of ways = 11P4

If all 4 items arranged in same spot then, no. of ways = 4! * 11

I'm just confused how to proceed further like if 2 items in 1 spot and other two in different spots... etc. Is my method of solving this is correct?

msk3002
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    Does the arrangements of items on a spot make a difference? – DJohnM Aug 30 '17 at 15:53
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    Are the four items distinct or identical? Does arrangement "within" a spot matter? If order within a spot doesn't matter, shouldn't it have been just $11$ ways, not $11\cdot 4!$ ways for all to occupy the same space? All items distinct and order within a spot not mattering, apply multiplication principle. There are $11$ available choices for where the first goes, $11$ choices for where the second goes, $11$ choices for where the third goes, etc... giving a total of... – JMoravitz Aug 30 '17 at 15:59
  • If arrangement within a spot matters, then consider rearrangements of the string 1234xxxxxxxxxx and how those correspond to arrangements you are wishing to count. E.g. 1x324xxxxxxxxx might correspond to the first item being in the first spot, and the remaining three items being in the second spot in the order $324$. How many rearrangements exist for this? – JMoravitz Aug 30 '17 at 16:01

1 Answers1

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What you've done so far is correct (assuming that the 4 items a distinct), and you could also work out the three remaining cases and add them all together. You did the 1+1+1+1 and single stack of 4 cases, the other cases are stacked as 3+1, 2+2, and 2+1+1.

There is however an easier way:

There are 11 places where you can put the first item.

The second item now has 12 locations where it can be placed - 11 spots on the ground (one of which is underneath the first item) or on top of the first item.

The third item now has 13 locations where it can be placed - 11 spots on the ground, directly on top of the first item, or directly on top of the second item.

etc.

So there are 11*12*13*14 = 24024 ways.

Let's verify this by doing it the long way and see if it gives the same answer.

  • 1+1+1+1 case: $^{11}P_4 = 7920$

  • 2+1+1 case: $^4C_2\ 2!\ ^{11}P_3 = 6*2*990 =11880$
  • - 2+2 case: $^4C_2/2\ 2!\ 2!\ ^{11}P_2 = 3*2*2*110 =1320$
  • - 3+1 case: $^4C_3\ 3!\ ^{11}P_2 = 4*6*110 =2640$
  • 4 case: $4! \ ^{11}P_1 = 24*11 =264$

Total:

$7920+11880+1320+2640+264 = 24024$

Jaap Scherphuis
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