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Combinatorics questions
Count without listing. Multiply choices when they happen in sequence, and decide whether order matters before reaching for a formula.
6 original questions · 20 marks · the combinatorics notes · Further Pure 2
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
From 10 different letters, how many ordered strings of 4 can be made, and how many unordered selections of 4?
Worked answer
Ordered: 10 × 9 × 8 × 7 = 5040, which is 10P4. Unordered: divide by the 4! = 24 orderings of each selection, giving 210 = 10C4. M1 A1 for the ordered count, A1 for the unordered count. The two always differ by exactly r!.In how many ways can 8 people be seated round a circular table, if only their positions relative to each other matter?
Worked answer
Fix one person to remove the rotational freedom, then arrange the other 7 in the remaining seats: 7! = 5040 ways. M1 for fixing one person, A1 for 5040. Counting 8! would treat each arrangement as 8 different ones, once for each rotation.How many distinct arrangements are there of the letters of STATISTICS?
Worked answer
There are 10 letters, with S appearing 3 times, T 3 times and I twice; A and C appear once each. Dividing the 10! orderings by the indistinguishable rearrangements gives 10!/(3! 3! 2!) = 3628800/72 = 50400. M1 for 10! over the repeats, A1 for the divisor, A1 for the total.A committee of 5 is chosen from 5 women and 7 men. How many committees contain at least 2 women?
Worked answer
Total committees: 12C5 = 792. Count the complement instead. No women: 7C5 = 21. Exactly one woman: 5 × 7C4 = 5 × 35 = 175. So the answer is 792 − 21 − 175 = 596. M1 for a complete method, A1 for the no-women count, A1 for the one-woman count, A1 for 596. Adding the cases with 2, 3, 4 and 5 women gives the same total in four times the work.How many integers from 1 to 500 inclusive are divisible by 3 or by 5?
Worked answer
Multiples of 3: 166. Multiples of 5: 100. Multiples of 15, counted in both: 33. By inclusion and exclusion the answer is 166 + 100 − 33 = 233. M1 for the three counts, A1 for subtracting the overlap, A1 for 233. Forgetting the overlap gives 266 and counts every multiple of 15 twice.Show that a set of 10 elements has exactly 512 subsets with an even number of elements.
Worked answer
The total number of subsets is 2¹⁰ = 1024, since each element is independently in or out. Expanding (1 − 1)¹⁰ by the binomial theorem gives the alternating sum of the 10Cr, which is 0, so the even-sized and odd-sized subsets are equal in number. Each group therefore has 1024/2 = 512 members. B1 for the total number of subsets, M1 for the binomial expansion, A1 for the alternating sum being zero, M1 for concluding that the two groups are equal, A1 for the printed result.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise combinatorics one question at a time
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