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$nHr = \frac{(n+r-1)!}{r!(n-1)!}$, for choosing $r$ elements from a set of $n$ elements allowing for repetition. Example: 3H2 = $\frac{(3+2-1)!}{2!(3-1)!} = 6$

If one element must (or must not) be included in every combination, use $_{n-1}C_{r-1}$ (or $_{n-1}C_r$). Example (must include): For $n=6$, $r=3$ and one element must be included, $_{6-1}C_{3-1} = 10$

The number of ways to choose $r$ from $n$ is equal to choosing $n-r$ from $n$: $_nC_r = _nC_{n-r}$. Example: $7C2 = 7C5 = 21$

$nCr = \frac{n!}{r!(n-r)!}$, where $n!$ is the factorial of $n$. Example: For $n=5$ and $r=2$, $5C2 = \frac{5!}{2!(5-2)!} = 10$

The $k$th coefficient in the expansion of $(a+b)^n$ is given by $_nC_k = \frac{n!}{k!(n-k)!}$. Example: The 3rd coefficient of $(a+b)^5$ is $5C2 = 10$

Given a multiset with $n$ types of items, the formula to find combinations is $\frac{(n_1+n_2+...+n_k)!}{n_1!n_2!...n_k!}$. Example: For a multiset with 3 As and 4 Bs, combinations of 7 items are $\frac{7!}{3!4!} = 35$.

Pascal's Identity is given by $_nC_r = _{n-1}C_{r-1} +_{n-1}C_r$. Example: $5C3 = 4C2 + 4C3 = 6 + 4 = 10$

To combine two distinct sets with $n$ and $m$ elements, choose $r$ from $n$, and $s$ from $m$, then multiply: $_nC_r × _mC_s$. Example: Choose 2 from 5 and 3 from 4, $5C2 × 4C3 = 10 × 4 = 40$ combinations.

When pairing elements, use $_{n}C_{2}$ to find the number of unique pairs. Example: For $n=4$ elements, $4C2 = \frac{4!}{2!(4-2)!} = 6$ pairs