Permutations and combinations questions, with answers

The questions, with answers

1. 1.How do you decide whether a question needs a permutation or a combination?

   In short: Ask whether swapping two chosen items gives a different outcome: if it does, order matters and you count permutations; if not, you count combinations.

   Choosing 3 people from 8 for a committee is a combination: the same three people are the same committee in any order, so the count is 8C3 = 8 × 7 × 6 / (3 × 2 × 1) = 56. Choosing a president, a secretary and a treasurer from the same 8 is a permutation, because swapping two people swaps their posts: 8P3 = 8 × 7 × 6 = 336. The two are linked — 336 = 56 × 3!, each committee arranged into posts in 3! ways — and that link is the safest way to remember the formulas. Words such as arrange, line up, rank and form numbers signal order; choose, select, team and committee usually do not.

2. 2.What are the counting rules behind every permutations and combinations question?

   In short: Multiply the choices when they happen together — this and then that — and add them when they are alternatives — this or that.

   With 4 shirts, 3 pairs of trousers and 2 pairs of shoes, an outfit is a shirt and trousers and shoes: 4 × 3 × 2 = 24 outfits. Travelling between two cities by one of 3 buses or one of 2 trains is 3 + 2 = 5 ways. Every formula in the topic is these two rules applied repeatedly. Three-digit numbers from the digits 1 to 6 with no digit repeated: 6 choices for the first place, 5 for the second and 4 for the third, so 6 × 5 × 4 = 120. If digits may repeat, each place has all 6 choices: 6³ = 216. Say the choices out loud place by place and the multiplication writes itself.

3. 3.How do you count the arrangements of a word with repeated letters?

   In short: Divide n! by the factorial of each letter's repeat count, because swapping two identical letters produces no new arrangement.

   ACCESS has 6 letters: A once, C twice, E once and S twice. If every letter were distinct there would be 6! = 720 arrangements, but each real arrangement has been counted 2! times for the two Cs and 2! times for the two Ss. So the answer is 720 / (2! × 2!) = 720 / 4 = 180. Count the repeats carefully before dividing — misreading a letter's count is the usual error, not the formula. A letter that appears three times divides by 3! = 6, not by 3. The same reasoning covers identical objects of any kind: 3 red and 2 blue flags in a row can be arranged in 5! / (3! × 2!) = 10 ways.

4. 4.How do you count arrangements where some items must stay together or apart?

   In short: Glue the items that must be together into one block, arrange the blocks, then multiply by the arrangements inside the block; 'never together' is the total minus 'together'.

   Arrange the letters of PENCIL so that the vowels E and I are always together. Treat EI as one block; with P, N, C and L that makes 5 units, arranged in 5! = 120 ways, and the block itself can be EI or IE, so 120 × 2 = 240. For the vowels never together, subtract from all arrangements: 6! − 240 = 720 − 240 = 480. The same idea seats people: 5 friends in a row with two particular friends side by side can sit in 4! × 2! = 48 ways. The mistake to avoid is forgetting the arrangements inside the block, which halves the answer — and the options usually include that half.

5. 5.How do you solve committee questions with 'at least' or 'at most'?

   In short: Either add the exact cases one by one or subtract the unwanted cases from the total — whichever leaves fewer cases to count.

   A committee of 4 is chosen from 6 men and 5 women and must include at least one woman. The unwanted case is a committee with no women: 6C4 = 15. The total is 11C4 = 330, so the answer is 330 − 15 = 315. The direct method agrees but takes four cases — one woman: 5 × 20 = 100; two: 10 × 15 = 150; three: 10 × 6 = 60; four: 5 × 1 = 5; total 315. The classic wrong method picks one woman first (5 ways) and then any 3 of the remaining 10 people (120 ways), getting 600, because it counts every committee with two or more women more than once. 'At least one' almost always means total minus none.

6. 6.How do you count arrangements around a circle?

   In short: Fix one person's seat to remove rotations: n people around a table can sit in (n − 1)! ways, and a necklace, which can be turned over, in (n − 1)!/2.

   Around a round table only the order of people relative to each other matters, so moving everyone one seat along gives the same seating. Fix one person and arrange the rest: 6 people can sit in 5! = 120 ways. A necklace of 6 different beads can also be turned over, which makes the clockwise and anticlockwise orders the same, so it has 120 / 2 = 60 arrangements. Conditions work as they do in a row: if two particular people must sit together at a table of 6, treat them as one unit, giving 5 units around the table in 4! = 24 ways, times 2 for the pair's own order — 48 seatings.

7. 7.How do you count numbers formed from digits when there are conditions?

   In short: Fill the most restricted place first — the last digit for 'even' or 'divisible by 5', the first digit when 0 is among the digits — and then the rest.

   How many three-digit even numbers can be formed from 1, 2, 3, 4 and 5 with no digit repeated? The last digit must be 2 or 4: 2 ways. The first place then has 4 remaining digits and the middle place 3, so 2 × 4 × 3 = 24. With 0 available, the first place is the restricted one: three-digit numbers from 0, 1, 2, 3 and 4 without repetition have 4 choices for the first digit (anything but 0), then 4 for the second, since 0 is now allowed, and 3 for the third: 48. Filling the places left to right without looking at the restriction is the source of most wrong answers here, including counting numbers that start with 0.

8. 8.How do permutations and combinations feed into probability questions?

   In short: Probability is favourable outcomes over all outcomes, and both are usually combination counts — count the favourable selections and divide by every selection.

   A committee of 3 is chosen at random from 4 men and 3 women. There are 7C3 = 35 possible committees. Committees of men only number 4C3 = 4, so the probability of an all-male committee is 4 / 35. The probability of exactly two women is 3C2 × 4C1 / 35 = 3 × 4 / 35 = 12 / 35. Use combinations for both the top and the bottom of the fraction, or permutations for both; an ordered count over an unordered one makes the answer 3! times too big here. The probability page works through the same idea with dice and with balls drawn from a bag.

How the diagnostic asks it

One question from the Aptitude bank, exactly as a sitting would show it. The bank has 8 on permutations & combinations and 84 across Aptitude.