"Permutations and Combinations" — Practice

Question 1

How many arrangements are there of the letters of the word ARRANGEMENT?

Accepted Answer

The word has eleven letters. Counting repeats: A appears twice, R twice, N twice, E twice, and G, M and T once each. Build as if all eleven letters were distinct, giving 11 factorial, which is 39916800. Then divide out the swaps you cannot detect, one factorial of 2 for each of the four repeated letters, giving a divisor of 2 times 2 times 2 times 2, which is 16. The number of distinct arrangements is 39916800 divided by 16, which is 2494800.

Question 2

In how many arrangements of the letters of MATHEMATICS do all four vowels occur together?

Accepted Answer

The vowels are A, A, E and I. Glue them into a single block. The remaining letters are the seven consonants M, M, T, T, H, C and S, so together with the block there are eight units to arrange. Two Ms and two Ts are still repeated, so the arrangements of the units number 8 factorial divided by 2 factorial twice, which is 40320 over 4, that is 10080. Inside the block the four vowels arrange among themselves with A repeated, giving 4 factorial over 2 factorial, which is 12. Both decisions are made, so multiply: 10080 times 12 gives 120960.

Question 3

Find the rank of the word MOTHER when all arrangements of its letters are listed in dictionary order.

Accepted Answer

The letters in alphabetical order are E, H, M, O, R, T, all distinct. At position one the letter used is M, and the available earlier letters are E and H, contributing 2 times 5 factorial, which is 240. At position two the letter is O with E, H, O, R, T available; earlier letters are E and H, contributing 2 times 4 factorial, which is 48. At position three the letter is T with E, H, R, T available; earlier are E, H and R, contributing 3 times 3 factorial, which is 18. At position four the letter is H with E, H, R available; earlier is E alone, contributing 1 times 2 factorial, which is 2. Positions five and six contribute nothing. The words before MOTHER number 240 plus 48 plus 18 plus 2, which is 308, so the rank is 309.

Question 4

How many four-digit numbers can be formed from the digits 1 to 7 if repetition is allowed, and how many if it is not?

Accepted Answer

With repetition allowed nothing is used up, so each of the four positions independently has all seven digits available. The count is 7 to the fourth power, which is 2401. Without repetition each digit chosen is removed from the pool, so the first position has 7 choices, the second 6, the third 5 and the fourth 4. The count is 7 times 6 times 5 times 4, which is 840. This is the permutation count P(7,4), equal to 7 factorial divided by 3 factorial.

Question 5

From 8 men and 5 women, how many committees of 4 can be formed containing at least 2 women?

Accepted Answer

At least two women means two, three or four women, and these cases are mutually exclusive, so count each and add. Two women and two men gives C(5,2) times C(8,2), which is 10 times 28, that is 280. Three women and one man gives C(5,3) times C(8,1), which is 10 times 8, that is 80. Four women gives C(5,4), which is 5. The total is 280 plus 80 plus 5, which is 365 committees. Note the structure: multiply within a case because both selections are made, and add across cases because exactly one case occurs.

Question 6

How many five-card hands from a standard 52-card deck contain at least one king? Explain why the complement is the better route.

Accepted Answer

The complement is better because counting directly requires four separate cases, for exactly one, two, three and four kings, while the complement requires one subtraction. The total number of hands is C(52,5), which is 2598960. Hands containing no king are drawn entirely from the 48 non-kings, giving C(48,5), which is 1712304. Subtracting gives 2598960 minus 1712304, which is 886656 hands with at least one king. The general pattern is that at least one is the complement of none, and none is almost always a single clean count.

Question 7

Six boys and four girls stand in a row. In how many ways can they be arranged so that no two girls are adjacent?

Accepted Answer

Complementary counting is awkward because several pairs of girls could be adjacent and the cases overlap, so use the gap method. Arrange the six boys first, in 6 factorial, which is 720 ways. Six boys in a row create seven gaps, one before each boy and one after the last, and placing the girls into distinct gaps guarantees that no two are adjacent. Choose four of the seven gaps and arrange the four girls in them, which is the ordered count P(7,4), equal to 7 times 6 times 5 times 4, that is 840. Both decisions are made, so the total is 720 times 840, which is 604800.

Question 8

Find the number of divisors of 360 and the sum of those divisors.

Accepted Answer

First factorise: 360 equals 2 cubed times 3 squared times 5 to the first power. A divisor is built by independently choosing an exponent for each prime, anywhere from zero up to the exponent available. That gives 4 choices for the power of 2, 3 for the power of 3 and 2 for the power of 5, so the number of divisors is 4 times 3 times 2, which is 24. For the sum, multiply out the sums of the available powers: (1 plus 2 plus 4 plus 8) times (1 plus 3 plus 9) times (1 plus 5), which is 15 times 13 times 6, giving 1170.

Question 9

Eight people sit around a round table. In how many ways can they be seated if two particular people must sit together, and in how many if they must not?

Accepted Answer

For the together count, glue the two people into a single block, leaving seven units around the table. Circular arrangements of seven units number 7 minus 1 factorial, which is 6 factorial, that is 720. Inside the block the two can swap, giving a factor of 2, so the together count is 1440. For the not-together count, take the complement. The unrestricted total is 8 minus 1 factorial, which is 5040. Subtracting gives 5040 minus 1440, which is 3600 seatings. The complement is the right route here because the constraint involves only a single pair.

Question 10

In how many ways can 12 distinct objects be divided into three groups of four, if the groups are labelled, and if they are unlabelled?

Accepted Answer

For labelled groups, choose the members of each group in turn and multiply, which collapses to 12 factorial divided by the product of three 4 factorials. That is 479001600 divided by 13824, which is 34650. For unlabelled groups, this overcounts, because swapping two whole groups of the same size produces the same division and has been counted as different. There are three equal-sized groups, so divide by 3 factorial, giving 34650 over 6, which is 5775. Note that the correction applies only because the groups have equal size; groups of different sizes could never be confused.

Question 11

Find the number of non-negative integer solutions of x + y + z + w = 10.

Accepted Answer

Picture ten identical items to be shared among four variables. Inserting three dividers among the items determines exactly one solution, since the counts between consecutive dividers give the values of the variables. So we are arranging ten items and three dividers in a row, which is thirteen positions of which three must be chosen for dividers. The count is C(13,3), which is 13 times 12 times 11 divided by 6, that is 286. Because the items are identical, we are choosing divider positions rather than arranging distinguishable objects, which is why the answer is a combination.

Question 12

Evaluate C(20,17) efficiently, and state how many subsets of a set of 6 elements contain at least one element.

Accepted Answer

For the first, use the symmetry identity, since choosing which 17 elements are in is the same as choosing which 3 are out. So C(20,17) equals C(20,3), which is 20 times 19 times 18 divided by 6, that is 1140. Computing the original form directly would require far larger factorials for the same answer. For the second, each of the 6 elements is independently in or out of a given subset, so there are 2 to the sixth power, which is 64, subsets in total. Removing the single empty subset leaves 63 subsets containing at least one element.

Practice — "Permutations and Combinations"

Sign in to save attempts to your dashboard and track your accuracy over time

Find a verified tutor

Join a live course

Browse coaching centres

About

Categories

Support

Documentation

Get Study Tips & Updates

Download Our App

Download for iOS

Download for Android

Popular Cities