Problems:
1. How many ways can you place 8 identical balls in 5 distinguishable buckets
(where buckets can be left empty)?
2. Three men and three women are assigned different numbers selected at
random from the integers 1 through 9. What is the probability the three numbers
assigned to the men are all greater than the three numbers assigned to the
women? (From the qualifying test!)
Solutions:
1. The 12-Fold Way of Combinatorics table says the answer to this question is
(8+5−1
) = (12
) = 495 ways. But… where did it pull that one from?
5−1
4
A useful way to visualize this problem is to use what some call the “Stars and
Bars” approach. First, imagine that the walls of the buckets have fused together
into one big, segmented container, leaving us with something like this:
Now, we can think of the balls and buckets instead as a linear arrangement of 8
balls (marked as stars) and 4 dividers.
(Two dividers with no stars in between indicate an empty bucket)
The outer walls will stay where they are, since no balls can go outside the buckets.
However, each different ordering of the 8 stars and 4 bars between them
represents a different way of placing the balls in the buckets.
So, how many of these orderings are there? We know that, no matter what, there
will be 12 total objects in a line between the outer walls. To make an
arrangement, all we have to do is choose which 4 of them we want to make
dividers, leaving the rest to be balls.
12
How many ways can we choose 4 out of 12 objects? ( 4 ) [“12 choose 4”] ways!
𝑥+𝑏−1
That’s where we get the formula ( 𝑏−1 ) from: when dealing with b
distinguishable buckets and x indistinguishable balls (and buckets can be left
empty), each combination is determined by taking the (𝒙 + 𝒃 − 𝟏) total objects
in between the “outer walls” and choosing (𝒃 − 𝟏) of them to be dividers, which
can be done “(𝑥 + 𝑏 − 1) choose (𝑏 − 1)” ways.
2. 𝑃𝑟𝑜𝑏𝑎𝑏𝑖𝑙𝑖𝑡𝑦 =
(# 𝑓𝑖𝑡𝑡𝑖𝑛𝑔 𝑎𝑟𝑟𝑎𝑛𝑔𝑒𝑚𝑒𝑛𝑡𝑠)
(# 𝑤𝑎𝑦𝑠 𝑡𝑜 𝑎𝑠𝑠𝑖𝑔𝑛 𝑡ℎ𝑒 6 𝑛𝑢𝑚𝑏𝑒𝑟𝑠)
Six different numbers out of the range 1-9 must be chosen. Then, the first person
gets one of the 6, the second has a choice of the remaining 5, the next has 4
possible numbers, etc, giving us 6*5*4*3*2*1 = 6! total possible arrangements.
Now, for the assigned numbers to fit the requirement given by the problem, the
highest 3 must go to the three men and the lowest 3 to the three women.
Assuming this happens, there are 3! orderings of the top 3 numbers and 3!
orderings of the bottom 3, giving us 3!*3! fitting arrangements
Therefore, the final probability is
3!∗3!
6!
, or
𝟏
𝟐𝟎
.