Mixed and applied problems

The hardest combinatorics questions combine permutations and combinations in a single problem. Mastery comes from a structured approach: first decide what must be selected, then how it must be arranged, and finally use the multiplication principle to combine.

A four-step strategy

For any combinatorial word problem:

1. Read carefully. Identify the objects and the constraints.
2. Decompose. Break the task into independent stages of selection and arrangement.
3. Apply formulas. Use $\binom{n}{r}$ for selections and $P(n, r)$ or factorials for arrangements.
4. Multiply (for "and") or add (for disjoint "or") the stage results.

Common problem types

Word problems with letter constraints

• All vowels together: glue them.
• No two vowels together: place consonants first; insert vowels in the gaps.

Committee / team problems

• "At least $k$ of type X": sum over admissible counts, or use complement.
• "At most $k$": similar.
• "Exactly $k$": one specific count.

Geometry problems

• Lines through pairs of points: $\binom{n}{2}$.
• Triangles from $n$ points (no three collinear): $\binom{n}{3}$.
• Diagonals of an $n$-gon: $\binom{n}{2} - n$.

Number-formation problems

• $r$-digit numbers from $n$ digits: $P(n, r)$ if no repetition (and possible leading-zero adjustment).
• Even / odd numbers: fix the last digit first.
• Divisible by $5$: last digit must be $0$ or $5$.

Worked examples

Example 1. From a group of $4$ men and $5$ women, in how many ways can a committee of $3$ men and $2$ women be formed if a particular woman, $W_1$, must be on the committee?

$W_1$ is fixed. Choose $1$ more woman from the remaining $4$: $\binom{4}{1} = 4$. Choose $3$ men from $4$: $\binom{4}{3} = 4$. Total: $4 \cdot 4 = 16$.

Example 2. How many words can be formed from the letters of "ARRANGE" so that the two $R$'s do not come together?

Letters: $A, R, R, A, N, G, E$ , $7$ letters with $A$ twice and $R$ twice.

Total arrangements: $\dfrac{7!}{2! \cdot 2!} = 1260$.

Arrangements with both $R$'s together: glue $RR$ , arrange $A, R\!R, A, N, G, E$, which is $6$ objects with $A$ twice: $\dfrac{6!}{2!} = 360$.

Required: $1260 - 360 = 900$.

Example 3. A bag has $4$ red and $6$ green balls. In how many ways can $4$ balls be drawn so that (a) all are red, (b) all are green, (c) two of each colour, (d) at least one is red?

(a) $\binom{4}{4} = 1$.

(b) $\binom{6}{4} = 15$.

(c) $\binom{4}{2} \cdot \binom{6}{2} = 6 \cdot 15 = 90$.

(d) Total $4$-draws: $\binom{10}{4} = 210$. None red: $\binom{6}{4} = 15$. At least one red: $210 - 15 = 195$.

Example 4. From $10$ points in a plane, no three collinear, find the number of (a) lines, (b) triangles.

(a) $\binom{10}{2} = 45$ lines.

(b) $\binom{10}{3} = 120$ triangles.

Example 5 (harder). How many ways can $5$ boys and $4$ girls be arranged in a row so that no two girls are adjacent?

Place the $5$ boys first: $5! = 120$ ways. Boys create $6$ gaps (before, between, and after them: $\_B\_B\_B\_B\_B\_$). Choose $4$ of these $6$ gaps for the girls: $\binom{6}{4} = 15$. Arrange the $4$ girls in those gaps: $4! = 24$.

Total: $120 \cdot 15 \cdot 24 = 43{,}200$.

Try it yourself

1. How many ways can $11$ players for a cricket team be chosen from $14$ if the captain is fixed?
2. In a group of $20$, in how many ways can a President, Vice-President and Secretary be chosen (distinct roles)?
3. How many words with all letters distinct can be formed from $A, B, C, D, E, F$ if no word begins with $A$?
4. How many committees of $5$ from $6$ men and $5$ women contain (a) exactly $3$ men, (b) at least $3$ men, (c) at most $2$ men?
5. How many four-digit even numbers can be made from $1, 2, 3, 4, 5, 6, 7$ with no repetition?
6. How many ways can the letters of "PERMUTATIONS" be arranged so that (a) all vowels are together, (b) the relative order of vowels and consonants is unchanged?
7. From $12$ points in a plane, $5$ of which are collinear, find the number of lines.
8. How many triangles are formed in the situation of (7)?
9. In how many ways can $4$ identical red and $3$ identical blue balls be arranged in a row?
10. How many $6$-digit numbers are even and have all distinct digits from $0$–$9$?
11. From $4$ Indian, $5$ American and $6$ Chinese members, how many delegations of $4$ have at least one Indian?
12. Find $n$ such that $\binom{n}{4} = \binom{n}{6}$.

Pitfalls / Tricks

• Always classify: selection vs arrangement. Many wrong answers come from picking the wrong type of formula.
• "At least one" = total minus "none".
• For arrangements where some items must not be adjacent, place the others first and insert into gaps.
• Insight. When stuck, write out the case structure first. For instance "at least $3$ men in committee of $5$ from $6+5$" means: $3$M+$2$W, $4$M+$1$W, $5$M+$0$W. Sum the three.