# Notation for combinations and permutations?

The number of ways of obtaining an ordered subset of $k$ elements from a set of $n$ elements is given by:

$$_nP_k = \frac{n!}{(n - k)!}$$

In other words, to get the permutations we form the fraction with denominator $(n - k)!$, only to get the last $k$ terms of the factorial in the numerator, i.e. n(n-1)(n-2)...(n-k+1).

Why not simply writing something like:

$$_nP_k = \prod_{i = n-k+1}^n i$$

Or some similar notation, that indicates (partial) product of the last k terms ?

Then, k combinations out of n elements would be:

$$_nC_k = \frac{\prod_{i = n-k+1}^n i}{\prod_{i = 1}^k i}$$

I guess, the question is why are we using fractions and factorials instead of notation denoting the last k terms of a factorial, such as products or similar?

Examples:

1. If three alphabets are to be chosen from A, B, C, D and E such that repetition is not allowed then in how many ways it can be done?

From $$_nP_k = \frac{n!}{(n-k)!}$$ for $n = 5$ and $k = 3$, we get

$$_5P_3 = \frac{5!}{(5-3)!} = \frac{1*2*3*4*5}{1*2} = 3*4*5 = 60$$

From $$_nP_k = \prod_{i = n-k+1}^n i$$ for $n = 5$ and $k = 3$, we get

$$_5P_3 = \prod_{i =5-3+1}^n i = 3*4*5= 60$$

2.In how many ways could 3 billiard balls be chosen from the (total) 16 balls?

From $$_nC_k = \frac{n!}{k!(n - k)!}$$ for $n = 16$ and $k = 3$, we get

$$_{16}C_3 = \frac{16!}{3!(16 - 3)!} = \frac{16!}{3!(13)!} = \frac{14*15*16}{1*2*3} = \frac{3360}{6} = 560$$

From $$_nC_k = \frac{\prod_{i = n-k+1}^n i}{\prod_{i = 1}^k i}$$ for $n = 16$ and $k = 3$, we get

$$_{16}C_3 = \frac{\prod_{i = 14}^{16} i}{\prod_{i = 1}^3 i} = \frac{14*15*16}{1*2*3} = 560$$

• I think $\Pi$ and $n!$ may achieve the same thing. However, it is like using the right tool for the right job. – sku Mar 16 '18 at 16:12

$$_n C_m=\binom{n}m\quad\text{and}\quad _nP_m=n^{\underline m}$$
where the notation $\binom{n}m$ is the same used for binomial coefficients, and $$n^\underline m:=n\cdot (n-1)\cdots (n-m+1)$$ is named falling factorial. This is the notation that I generally use, and we also have that $$\binom{n}m=\frac{n!}{(n-m)!\,m!}=\frac{n^\underline m}{m!}$$