Permutation & Combination • Topic 2 of 4

Selections

A selection counts groups, not orders — picking {A, B, C} is the same outcome as {C, B, A}. The count is nCr = n!/(r!(n−r)!), the number of ways to choose r of n. Two facts save time in CAT. First, nCr = nC(n−r): choosing 8 of 10 to keep is the same as choosing 2 to leave out, so compute the easier side. Second, the sum nC0 + nC1 + … + nCn = 2ⁿ counts every possible subset, which gives "at least one" as 2ⁿ − 1 instantly. Selection problems often split into cases joined by "and" (multiply) and "or" (add): a committee of 2 men and 3 women from 5 men, 6 women is 5C2 × 6C3. When a problem says "at least", it is usually faster to count the complement and subtract from the total.

✅ Solved examples

1. In how many ways can a committee of 3 be chosen from 8 people?
8C3 = (8 × 7 × 6)/(3 × 2 × 1) = 336/6 = 56.
2. A team of 2 men and 3 women is formed from 5 men and 6 women. How many teams?
5C2 × 6C3 = 10 × 20 = 200.
3. From 12 players, how many ways to leave out 9 (i.e. pick 3)?
12C9 = 12C3 = (12 × 11 × 10)/6 = 220.
4. How many subsets (including empty and full) does a set of 5 elements have?
2⁵ = 32. With "at least one element" it would be 2⁵ − 1 = 31.

✏️ Practice — try these, take hints as needed

1. Choose 2 fruits from 7 distinct fruits. How many ways?
Order ignored.
7C2.
(7 × 6)/2.
21
2. A committee of 4 from 10 people. How many ways?
10C4.
(10 × 9 × 8 × 7)/(4!).
5040/24.
210
3. From 6 men and 4 women, select 3 men and 2 women.
Multiply the two choices.
6C3 × 4C2.
20 × 6.
120
4. How many ways to choose at least one book from 4 distinct books?
Total subsets 2⁴.
Remove the empty set.
16 − 1.
15
5. A student must answer 7 of 10 questions. How many choices of questions?
10C7 = 10C3.
(10 × 9 × 8)/6.
720/6.
120

📝 Topic test — 8 questions

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