Q.Find $n$ if ${}^{n-1}P_3 : {}^{n}P_4 = 1 : 9$.

The Intuition: Two Ways to Choose

Imagine you have a group of 10 friends, and you need to pick 3 of them to form a committee. One way to think about this is: you are choosing the 3 people who will be on the committee. But there is another, equally valid way to think about it: you are rejecting the 7 people who will not be on the committee.

Choosing 3 to include is the same decision as choosing 7 to exclude. Every time you pick a set of 3, you automatically determine the set of 7 who are left out. There is a perfect one-to-one match between the two choices.

This is the heart of the symmetry property: the number of ways to choose $k$ items from $n$ is exactly the same as the number of ways to choose $n-k$ items from $n$.

The Precise Statement

Where $\binom{n}{k}$ (read "n choose k") is the number of combinations — the number of distinct subsets of size $k$ you can pick from a set of $n$ distinct objects.

This holds for any non-negative integers $n$ and $k$ where $0 \leq k \leq n$.

Why It Works (The Algebraic Proof)

The formula for combinations is:

Now compute $\binom{n}{n-k}$:

The denominator is just $k!(n-k)!$ written in a different order. Since multiplication is commutative, the two expressions are identical.

Special Cases That Make Sense

• $k = 0$: $\binom{n}{0} = 1$ (there is exactly one way to choose nothing). By symmetry, $\binom{n}{n} = 1$ (one way to choose everything). Both make sense — you either take nothing or take all.

• $k = 1$: $\binom{n}{1} = n$. Symmetry gives $\binom{n}{n-1} = n$. Choosing 1 person to include is the same as choosing $n-1$ people to exclude — there are $n$ choices in either case. …