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AS & A Level · Practice questions · Permutations: arrangements in a line and with restrictions

Count it without listing.

Six original Cambridge-style questions on permutations in a line with repetition and restriction, at Paper 5 standard.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
The method

Restrictions first, then fill the rest.

01
Recall
[1 marks]

Find the number of different arrangements of the letters of the word LETTER.

  • 6 letters with E twice and T twice: \(\dfrac{6!}{2!\,2!} = 180\). ✓
02
Calculation
[3 marks]

Find the number of arrangements of the letters of STATISTICS (a) in total, (b) which begin and end with S.

  • (a) \(\dfrac{10!}{3!\,3!\,2!} = 50\,400\). ✓
  • (b) The middle 8 letters are S, T, T, T, A, I, I, C: \(\dfrac{8!}{3!\,2!} = 3360\). ✓
03
Calculation
[3 marks]

Eight people, including Ana, Ben and Cal, stand in a line. Find the number of arrangements in which Cal is at one end and Ana and Ben are not next to each other.

  • Cal at an end: 2 ways. The other 7 fill 7 places. ✓
  • Those 7 in total: \(7! = 5040\); with Ana and Ben together: \(6! \times 2 = 1440\). ✓
  • \(2 \times (5040 - 1440) = 7200\). ✓
04
Calculation
[3 marks]

Four-digit numbers are made from the digits 1, 2, 3, 4, 5, 6 with no digit repeated. How many of them are even and greater than 4000?

  • Case 1, first digit 4 or 6: 2 choices of first digit, 2 even digits left for the last place, \(4 \times 3\) for the middle: 48. ✓
  • Case 2, first digit 5: 3 even digits for the last place, \(4 \times 3\) for the middle: 36. ✓
  • Total \(48 + 36 = 84\). ✓
05
Analysis
[4 marks]

Find the number of arrangements of the letters of BANANAS (a) in total, (b) in which no two As are next to each other.

  • (a) 7 letters with A three times and N twice: \(\dfrac{7!}{3!\,2!} = 420\). ✓
  • (b) Arrange B, N, N, S: \(\dfrac{4!}{2!} = 12\) ways, creating 5 gaps. ✓
  • The three As are identical, so choose 3 of the 5 gaps: \(\dfrac{5 \times 4 \times 3}{3!} = 10\). ✓
  • \(12 \times 10 = 120\). ✓
06
Analysis
[5 marks]

Nine friends go to a concert and sit in two rows: 4 seats in the front row and 5 in the back row. Find the number of ways they can sit if

(a) there are no restrictions, (b) Dev and Eli must sit next to each other in the front row, (c) Dev and Eli must sit in different rows.

  • (a) \(9! = 362\,880\). ✓
  • (b) The front row has 3 pairs of adjacent seats, and Dev and Eli can sit in either order: \(3 \times 2 = 6\). The other 7 fill 7 seats: \(7! = 5040\). Total \(30\,240\). ✓
  • (c) Dev front and Eli back: \(4 \times 5 = 20\); or the other way round: 20. So 40 ways, then \(7!\) for the rest. ✓
  • \(40 \times 5040 = 201\,600\). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.