Exercise practice: Permutations and Combinations

4.1 How the exercises are laid out

Exercise 6.1 applies the multiplication principle to digits, codes, coins and flags. Exercise 6.2 is factorial arithmetic and small factorial equations. Exercise 6.3 counts arrangements, with and without repeated letters, and includes the hard MISSISSIPPI and PERMUTATIONS questions. Exercise 6.4 counts selections: chords, cards, committees and course choices. The miscellaneous exercise mixes arrangements of letters with committees and card selections.

This page works one representative question from each block with the full reasoning.

4.2 Exercise 6.1 and 6.2

Question 1 of 6.1 asks for three digit numbers from 1, 2, 3, 4 and 5. With repetition the count is 5 cubed, that is 125. Without repetition the first place has 5 choices, then 4, then 3, giving 60.

Question 4 asks for telephone numbers starting with 67 using digits 0 to 9 without repetition. The first two places are fixed, so three places remain and the digits 6 and 7 are already used. The count is 8 times 7 times 6 equal to 336.

Question 1 of 6.2 evaluates 8 factorial and 4 factorial minus 3 factorial. The first is 40320 and the second is 24 minus 6 equal to 18. Question 2 asks whether 3 factorial plus 4 factorial equals 7 factorial: the left side is 6 plus 24 equal to 30 while the right side is 5040, so the statement is false.

Question 4 of 6.2 solves 1 by 6 factorial plus 1 by 7 factorial equal to x by 8 factorial. Write 8 factorial as 8 times 7 times 6 factorial and multiply through by 8 factorial. The equation becomes 56 plus 8 equal to x, so x is 64.

4.3 Exercise 6.3

Question 1 asks for three digit numbers from 1 to 9 with no repetition. The first place has 9 choices, then 8, then 7, giving 9 times 8 times 7 equal to 504.

Question 4 asks for four digit numbers from 1 to 5 and how many are even. The total is 5P4 equal to 120. For an even number the last digit must be 2 or 4, so fix it in 2 ways and fill the first three places from the remaining four digits in 4P3 equal to 24 ways, giving 48 even numbers.

Question 10 works with MISSISSIPPI. The word has 11 letters with 4 I's, 4 S's and 2 P's, so the total distinct arrangements are 11 factorial divided by 4 factorial times 4 factorial times 2 factorial, equal to 34650. To count the arrangements where all four I's are together, treat the block of four I's as one object; the objects are then the block, 4 S's, 2 P's and the M, which are 8 objects with repeats of S and P, giving 8 factorial divided by 4 factorial times 2 factorial, equal to 840. The required count is 34650 minus 840 equal to 33810.

Question 11 arranges PERMUTATIONS. For part (i) with P at the start and S at the end, the remaining 10 letters include T twice and the rest different, so the count is 10 factorial by 2 factorial equal to 1814400. Part (ii) keeping the vowels together treats the five vowels as one block, and the count then divides by the internal arrangements of the block.

4.4 Exercise 6.4

Question 1 uses nC8 equal to nC2. The identity nCr equal to nC(n minus r) gives r equal to 2 on one side or 8 on the other, so n is 10. The question then asks for 10C2, which is 45.

Question 3 asks for chords through 21 points on a circle. Every chord is fixed by its two endpoints and the order of the endpoints does not matter, so the count is 21C2 equal to 210.

Question 4 selects 3 boys from 5 and 3 girls from 4. The choices are independent, so multiply: 5C3 times 4C3 equal to 10 times 4 equal to 40.

Question 6 asks for 5 card combinations with exactly one ace. Choose the ace in 4 ways and choose the remaining four cards from the 48 cards that are not aces, which gives 4 times 48C4 equal to 4 times 194580 equal to 778320.

Question 9 chooses 5 courses with 2 compulsory ones. The two compulsory courses take 2 places, so 3 places remain from the other 7 courses, giving 7C3 equal to 35.

4.5 Miscellaneous exercise

Question 1 asks for words with 2 vowels and 3 consonants from DAUGHTER, which has 3 vowels and 5 consonants. Choose the vowels in 3C2 equal to 3 ways and the consonants in 5C3 equal to 10 ways, then arrange the five chosen letters in 5 factorial equal to 120 ways. The total is 3 times 10 times 120 equal to 3600.

Question 3 forms a committee of 7 from 9 boys and 4 girls. For exactly 3 girls choose 3 girls from 4 and 4 boys from 9: 4C3 times 9C4 equal to 4 times 126 equal to 504. For at least 3 girls add the cases of 3 and 4 girls: 504 plus 4C4 times 9C3 equal to 504 plus 84 equal to 588. For at most 3 girls add the 0, 1, 2 and 3 girl cases, or subtract the 4 girl case from the total 13C7; both give the same answer.

Question 5 asks for 6 digit numbers from 0, 1, 3, 5, 7 and 9 divisible by 10 with no repetition. A number divisible by 10 must end in 0, so fix the last place and arrange the other five digits in 5 factorial equal to 120 ways; the first digit cannot be 0, but 0 is already used at the end, so all 120 arrangements are valid.

Question 11 arranges ASSASSINATION with all S's together. Treat the four S's as one block; the remaining letters are A three times, I twice, N twice and the block, which gives 10 objects with repeats. The count is 10 factorial divided by 3 factorial times 2 factorial times 2 factorial.

4.6 Common slips in these exercises

Three errors dominate. Using nPr when the question asks for a committee or a hand, where order is irrelevant. Forgetting the leading zero restriction in numbers formed from digits that include 0. Counting an at least case by choosing one item of the required kind first, which double counts combinations that contain more than one such item.

A fourth slip is subtracting the together case from the total when the together case itself was computed with the wrong number of objects. Count the block as one object and recount the repeated letters inside the remaining objects, then check that the two counts use the same denominator pattern.

A final check is to compute a small case by listing. If n equals 3 or 4, list the objects and confirm the formula answer. A listing check catches interpretation errors that arithmetic cannot.