A-Level Probability: worked solution

Question

Counting every 4-letter code that can be formed from the letters of BOOKKEEPER with no restriction, find the total number possible.

Step 1: sort out what letters are available. BOOKKEEPER = B, O, O, K, K, E, E, P, E, R So the letter counts are B x1, O x2, K x2, E x3, P x1, R x1 That is 6 DIFFERENT letters, of which O and K can be used twice and E up to three times. A code is an ordered string of 4 letters, so for each pattern of repeats we count which letters are chosen and then how many orders they give. Case 1: all four letters different. Choose 4 of the 6 different letters: 6C4 = 15. Arrange 4 different letters: 4! = 24. 15 x 24 = 360. Case 2: exactly one letter repeated twice, plus two other different letters. The repeated letter must have at least 2 copies, so it is O, K or E: 3 choices. Choose 2 more letters from the 5 remaining different letters: 5C2 = 10. Arrange 4 letters of which 2 are identical: 4!/2! = 12. 3 x 10 x 12 = 360. Case 3: two different letters, each used twice. Choose 2 of the three repeatable letters O, K, E: 3C2 = 3. Arrange 4 letters in two identical pairs: 4!/(2!2!) = 6. 3 x 6 = 18. Case 4: one letter used three times plus one other letter. Only E occurs three times, so the tripled letter is forced: 1 choice. Choose the remaining letter from the other 5 different letters: 5. Arrange 4 letters of which 3 are identical: 4!/3! = 4. 1 x 5 x 4 = 20. Case 5: all four letters the same is impossible, since no letter appears 4 times. Step 2: add the cases, which are mutually exclusive and cover every code. 360 + 360 + 18 + 20 = 758. So 758 different 4-letter codes can be formed.