# Arrangements of 3 baskets, 2 misses through Combinations or Permutations?

Need some help with a basic Permutations/Combinations question. In the linked Khan Academy video, Sal is evaluating possibilities regarding Baskets and Misses. The end goal (for the video) is the find the probability of getting atleast 3 baskets.

At a particular point in the video (1:00), Sal is evaluating the number of ways in which, out of 5 shots, we can have 3 Baskets and 2 misses. One way for example is BBBMM another way is MMBBB

it seems like order matters as BBBMM is not the same as MMBBB

Yet, Sal explicitly mentions using the "combinations" formula and logic to get this value, which appears to be correct as per the given number of combinations below

BBBMM BBMBM BBMMB BMBBM BMBMB BMMBB MBBBM MBBMB MBMBB MMBBB

I can reconcile this by thinking of it as Permutations with repetition leading to 5!/(3!2!) However, I would like to know what point Sal is trying to get across by presenting this as a Combinations exercise as opposed to permutations, and whether there's a general link between permutations with repetition and combinatorics. Thanks!

• It's combinations because you are choosing which of the shots are baskets. The order of these choices does not matter. E.g. the sequence MBMBB corresponds to choosing shots {2,4,5} to be baskets, which is the same as choosing {4,2,5} or {4,5,2}. – Mike Earnest May 7 '18 at 21:22
• Ahh, so that’s the way to think about it. So depending on the strategy we are taking, in some problems we can use both Permutations and Combinations. Which makes a lot of sense because simple permutations can actually be considered combinations with the an added factorial for order – user3233029 May 7 '18 at 21:45

I would like to know ... whether there's a general link between permutations with repetition and combinatorics.

There is an fundamental link between permutations (denoted $^nP_r$, used for counting ordered selections) and combinations ($^nC_r$, used for counting unordered selections), namely: $$^nC_r = \frac{^nP_r}{r!}$$

Of course, this is evident in the algebraic definitions: $$^nP_r = \frac{n!}{(n-r)!}, ^nC_r = \frac{n!}{r!(n-r)!}$$ More importantly, however, it is evident on an intuitive level: Combinations are just permutations where we don't care about order. Since there are $r!$ ways to order $r$ (distinct) objects, we should divide the amount of ordered selections ($^nP_r$) by $r!$ to get the amount of unordered selections ($^nC_r$) because ordered selections (permutations) over count by a factor of r! when compared unordered selections (combinations).

Example:

Say we have 4 items labeled 1, 2, 3, 4, and we want to select 3 of them (here n = 4, r = 3). We could write the possible ordered selections like this

$\begin{array}{c|cc} items & permutations \\ \hline 123 & (123), (132), (213), (231), (312), (321)\\ 124 & (124), (142), (214), (241), (412), (421)\\ 134 & (134), (143), (314), (341), (413), (431)\\ 234 & (234), (243), (324), (342), (423), (432)\\ \end{array}$

Thus, we would count $6\times4 = 24$ ordered selections (i.e. $^4P_3$). However, if we wanted unordered selections (which there are 4 of), we might notice that each unordered selection is over counted 6 (i.e. $3!$) times when counting ordered selections and thus, to count the amount of unordered selections, just divide the amount of ordered selections by 6. The reason we are dividing by 6 here is because there are 6 (i.e. 3!) ways to permute a set of 3 items. The key observation here is: The number of times each unordered set is repeated is the same as the number of distinct ways to permute that set.

By the same argument applied more generally: to calculate the number of unordered selections, you can divide the number of ordered selections by the amount of times each unordered selection is repeated (i.e. r!). Hence the relationship stated previously: $^nC_r = \frac{^nP_r}{r!}$

Permutations approach: $\frac{^5P_5}{(^3P_3)(^2P_2)} = \frac{5!}{(3!)(2!)} = 10$. $^5P_5$ orders the 5 objects (3 B's and 2 M's) and then $^3P_3$ and $^2P_2$ each remove the over counting of B's and M's (respectively) because of the repeated elements.
Combinations approach: $^5C_3 = 10$. Imagine there are 5 spaces for the letters to go. All that needs to be done is to choose 3 of those spaces for the B's to go, and then the M's will be forced to go in the remaining spaces. Note: you could also write this as $^5C_2$ by the same logic (here we see an intuitive example of the property $^nC_r ={^{n}C_{n-r}}$).