Why Choosing a Team Is Easier to Count Than Arranging One

Picking 3 students for a quiz team is different from giving them the posts of captain, deputy and reserve. In the first case order does not matter — that is a combination. Combinations count selections, and they appear in probability, the binomial theorem and everyday choices.

This lesson covers the nCr formula and its identities, selections from different items, selections from items that are not all different, and mixed problems.

**Each selection of r objects from n can be arranged in $r!$ ways, so ${}^n{P}_r={}^n{C}_r\times r!$ and ${}^n{C}_r=\frac{n!}{r!(n-r)!}$; choosing r objects is the same as rejecting the other $n-r$, so ${}^n{C}_r={}^n{C}_{n-r}$, and Pascal's rule states ${}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$.

Identities:**

- ${}^n{C}_0={}^n{C}_n=1$
- ${}^n{C}_r={}^n{C}_{n-r}$
- If ${}^n{C}_x={}^n{C}_y$, then $x=y$ or $x+y=n$
- Pascal's rule — ${}^n{C}_r+{}^n{C}_{r-1}={}^{n+1}{C}_r$

Worked example. ${}^{10}{C}_3=\frac{10\times 9\times 8}{3\times 2\times 1}=120$, and ${}^{10}{C}_7=120$ as well.

Worked example 2. If ${}^n{C}_8={}^n{C}_6$, then $n=8+6=14$, so ${}^{14}{C}_2=\frac{14\times 13}{2}=91$.

Pascal's rule check. ${}^5{C}_2+{}^5{C}_1=10+5=15={}^6{C}_2$.

An everyday example. Choosing 2 fillings from 5 for a dosa gives ${}^5{C}_2=10$ choices, and deciding which 3 to leave out gives the same 10.

The substance. Pascal's rule has a simple reason — a selection from $n+1$ items either includes one particular item or it does not.

**When all items are different, a selection is counted with ${}^n{C}_r$; 'must include' and 'must exclude' conditions reduce n and r first, 'at least' conditions are handled by cases or by subtracting from the total, and there are $2^n-1$ ways to choose at least one item from n.

Worked example. A committee of 5 is chosen from 6 boys and 4 girls.

- Any 5** — ${}^{10}{C}_5=252$
- Exactly 2 girls — ${}^4{C}_2\times {}^6{C}_3=6\times 20=120$
- At least 1 girl — $252-{}^6{C}_5=252-6=246$

Any number of items. Each of n items is either taken or left, so there are $2^n$ selections in all, or $2^n-1$ if at least one must be taken. From 5 different books, at least one can be chosen in $2^5-1=31$ ways.

An everyday example. Picking 3 elective subjects from a list of 8 can be done in ${}^8{C}_3=56$ ways.

The substance. 'At least one' is fastest by the complement — the total minus the selections with none.

**If a collection has p identical items of one kind, q of another and r of a third, the number of selections containing at least one item is $(p+1)(q+1)(r+1)-1$, and when n different items are also present the product is multiplied by $2^n$ before subtracting 1.

Why it works.** From p identical items you can take 0, 1, 2 and so on up to p — that is $p+1$ choices, not $2^p$, because identical items cannot be told apart. Subtracting 1 removes the empty selection.

Worked example. A basket holds 3 identical mangoes, 4 identical bananas and 2 identical apples. The number of ways to take at least one fruit is

$$(3+1)(4+1)(2+1)-1=60-1=59$$

If the basket also holds 2 different oranges, the count becomes $4\times 5\times 3\times 2^2-1=239$.

Worked example 2 (divisors). $360=2^3\times 3^2\times 5$ has $(3+1)(2+1)(1+1)=24$ divisors, because each divisor chooses how many of each prime factor to take.

An everyday example. Packing a tiffin from 5 identical rotis and 3 identical laddoos, with at least one item, gives $6\times 4-1=23$ possibilities.

The substance. **Identical items give $p+1$ choices, different items give $2^p$** — mixing the two is the classic slip.

Mixed problems are solved in two stages: first select the items with combinations, then arrange the selected items with permutations, and multiply the two counts.

Worked example. How many 5-letter words with 2 different vowels and 3 different consonants can be formed from 5 vowels and 7 consonants?

- Select the vowels: ${}^5{C}_2=10$
- Select the consonants: ${}^7{C}_3=35$
- Arrange the 5 chosen letters: $5!=120$

$$10\times 35\times 120=42000$$

Geometry link. 10 points with no three collinear determine ${}^{10}{C}_2=45$ lines and ${}^{10}{C}_3=120$ triangles, and a decagon has $45-10=35$ diagonals.

An everyday example. A housing society choosing a committee of 4 from 8 candidates and then naming a president and secretary from that committee has ${}^8{C}_4\times {}^4{P}_2=70\times 12=840$ possible outcomes.

The substance. Ask 'does order matter?' separately at each stage — selection and arrangement often occur in the same problem.