How it works
The number of ways to choose k objects from n when order does not matter is
C(n, k) = n! ÷ (k! (n − k)!)
A quick way to compute it: multiply k numbers counting down from n, then divide by k!. So C(10, 3) = 10 × 9 × 8 ÷ 6 = 120. Choosing who is in is the same as choosing who is out, so C(n, k) = C(n, n − k).
Required members: put them in first and choose the rest from who is left. Two separate groups (3 boys and 2 girls): multiply the choices for each group. "At least" conditions: add up the cases, or subtract the unwanted cases from the total.
Combinations practice problems with solutions
These problems come from Mathyessy's question bank. Try each one before opening the solution.
How many ways can you choose 2 toppings from 5 toppings?
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Answer: 10
Order does not matter: C(5, 2) = 5 × 4 ÷ 2 = 10.
A committee of 4 is chosen from 12 people. One specific person must be included. How many committees are possible?
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Answer: 165
Put the required person in. Then choose the other 3 from the remaining 11: C(11, 3) = 11 × 10 × 9 ÷ 6 = 165.
From 7 men and 6 women choose 5 people with exactly 3 men and 2 women. How many groups are possible?
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Answer: 525
Choose the men and women separately and multiply: C(7, 3) × C(6, 2) = 35 × 15 = 525.
From 6 men and 5 women choose 4 people with at least 2 women. How many groups are possible?
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Answer: 215
Split by the number of women. 2 women: C(5,2) × C(6,2) = 10 × 15 = 150. 3 women: C(5,3) × C(6,1) = 10 × 6 = 60. 4 women: C(5,4) = 5. Total 150 + 60 + 5 = 215.
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