4.5 Counting Principles: Permutations and Combinations

4.5 Counting Principles: Permutations and Combinations

Counting problems on an entrance test are rarely difficult arithmetic. They are difficult classification: almost every mark lost here comes from picking the wrong tool, not from multiplying badly. This section gives you the tools, then the decision rule that selects between them.

1. The Fundamental Counting Principle

The fundamental counting principle (FCP) says that if a task splits into independent stages with $m$, $n$ and $p$ choices, the number of complete outcomes is $m \times n \times p$. A canteen offering 4 kinds of rice, 5 ulam and 3 drinks sells $4 \times 5 \times 3 = 60$ distinct meal combinations.

The FCP also handles restrictions on the fly, one stage at a time. A 4-digit PIN with no repeated digit has $10 \times 9 \times 8 \times 7 = 5{,}040$ possibilities, because each stage removes one digit from the pool. If repetition were allowed, the count would be $10^4 = 10{,}000$ instead. Whenever repetition is permitted, the answer is a power, never a permutation.

2. Factorials and Linear Permutations

$n$ factorial is $n! = n(n-1)(n-2)\cdots(2)(1)$, with the convention $0! = 1$. It counts the arrangements of $n$ distinct objects in a row. Memorise the small values: $1, 2, 6, 24, 120, 720, 5040, 40320$ for $n = 1$ to $8$.

A permutation is an ordered selection. Choosing and ordering $r$ items from $n$ distinct items gives

$_nP_r = \frac{n!}{(n-r)!} = \underbrace{n(n-1)(n-2)\cdots}_{r \text{ factors}}$

Never expand the factorials in full — just multiply $r$ descending factors. For example $_{10}P_4 = 10 \times 9 \times 8 \times 7 = 5{,}040$.

Permutations with repeated items

When some of the $n$ objects are indistinguishable, divide out the arrangements you cannot tell apart. With $n_1$ copies of one type, $n_2$ of another, and so on:

$\text{distinct arrangements} = \frac{n!}{n_1!\,n_2!\cdots n_k!}$

The word BATANGAS has 8 letters, three of them A, so there are $8!/3! = 40{,}320/6 = 6{,}720$ distinct arrangements.

Circular and reflective arrangements

Around a circle only relative position matters, because rotating everyone one seat clockwise produces the same arrangement. Fix one person and arrange the rest:

$\text{circular permutations} = (n-1)!$

Eight friends around a round table can be seated in $7! = 5{,}040$ ways. If the arrangement can also be flipped over — keys on a key ring, beads on a necklace or bracelet — then each arrangement and its mirror image are the same physical object, so halve the count:

$\text{reflective circular} = \frac{(n-1)!}{2}$

Six distinct charms on a bracelet therefore admit $5!/2 = 120/2 = 60$ arrangements.

3. Combinations

A combination is an unordered selection:

$_nC_r = \frac{n!}{r!\,(n-r)!} = \frac{_nP_r}{r!}$

The division by $r!$ is the entire difference from a permutation: it erases the $r!$ orderings of each chosen group. Two properties save time on a calculator-free paper:

• Symmetry: $nC_r = {}nC{n-r}$. Choosing 7 of 10 is the same as rejecting 3 of 10, so ${10}C_7 = {}_{10}C_3 = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$.
• Edges: $_nC_0 = {}_nC_n = 1$ and $_nC_1 = n$.

4. The Decision Table

The fastest test: swap two of the chosen items. If you now have a genuinely different result, order matters (permutation); if it is the same result, it does not (combination). President-and-Secretary Ana-then-Ben differs from Ben-then-Ana, so that is a permutation. A committee of Ana and Ben is the same committee either way, so that is a combination.

5. Worked Example 1 — a committee with restrictions

Problem. A student council has 12 eligible members: 7 women (including Ana) and 5 men (including Ben). A 5-member committee is to be formed.

(a) How many committees are possible? Order is irrelevant: $_{12}C_5 = \frac{12 \times 11 \times 10 \times 9 \times 8}{5 \times 4 \times 3 \times 2 \times 1} = \frac{95{,}040}{120} = 792$

(b) Exactly 3 women and 2 men? Choose within each pool, then multiply by the FCP: $_7C_3 \times {}_5C_2 = 35 \times 10 = 350$

(c) At least 4 women? Split into cases and add: $({}_7C_4 \times {}_5C_1) + ({}_7C_5 \times {}_5C_0) = (35 \times 5) + (21 \times 1) = 175 + 21 = 196$

(d) The committee must include Ana. Ana's seat is fixed, so choose the other 4 from the remaining 11 people: $_{11}C_4 = \frac{11 \times 10 \times 9 \times 8}{24} = \frac{7{,}920}{24} = 330$

(e) Ana and Ben cannot both serve. Count the forbidden committees and subtract. If both are in, the other 3 come from the remaining 10 people: $_{10}C_3 = 120$. So the allowed total is $792 - 120 = 672$

Restriction problems almost always yield to one of three moves: fix the required member, split into cases, or subtract the forbidden count from the total. The subtraction route is usually the shortest when the words "not", "at least" or "cannot both" appear.

6. Worked Example 2 — arrangements of a word

Problem. Using all the letters of MANILA, find (a) the number of distinct arrangements, (b) the number in which the two A's are together, and (c) the number in which the two A's are not together.

(a) MANILA has 6 letters: M, A, N, I, L, A. Two of them are A; all the others are unique. $\frac{6!}{2!} = \frac{720}{2} = 360$

(b) Block method. Glue the two A's into a single unit "AA". You now arrange 5 objects — AA, M, N, I, L — in a row: $5! = 120$ There is no extra internal factor here because the two A's are identical; had the block held two different letters, you would multiply by $2!$ for their internal order.

(c) By subtraction. $360 - 120 = 240$

Gap-method check. Arrange the four non-A letters first: $4! = 24$ ways. That creates 5 gaps (one before each letter, one after the last):

  _ M _ N _ I _ L _        <- 5 gaps, choose 2 for the A's

Choose 2 of the 5 gaps for the A's, order irrelevant since the A's are identical: $_5C_2 = 10$. Total $24 \times 10 = 240$, matching the subtraction. Two independent methods agreeing is the best confirmation available on a timed paper.

7. Common Traps

1. Using a permutation for a committee. Handshakes among 10 people are ${10}C_2 = 45$, not ${10}P_2 = 90$; the permutation counts each handshake twice.
2. Forgetting to halve for a key ring. Seven keys on a ring give $6!/2 = 360$, not $6! = 720$.
3. Losing the internal order of a block. If two distinct people must sit together, multiply by 2 after treating them as one unit.
4. Forgetting the circular reduction inside a block problem. Six friends at a round table with two who must sit together gives $(5-1)! \times 2 = 48$, not $5! \times 2 = 240$.
5. Attacking "not adjacent" head-on. Count the total, count the together-cases, subtract.
6. Ignoring repeated letters. Dividing by the factorials of the repeat counts is not optional.
7. Using $_nP_r$ when repetition is allowed. Repetition means $n^r$; the descending-factor formula assumes each item is used once.
8. Adding cases that overlap. When splitting "at least 4 women" into cases, the cases must be mutually exclusive by construction — 4 women and 5 women cannot both describe the same committee.