Permutations of Distinct Objects (nPr) for CA Foundation

Understand Permutations of Distinct Objects

Key formulas to remember

Permutation of r out of n distinct objects

nPr = n! ÷ (n − r)! = n(n − 1)(n − 2)...(n − r + 1)

Valid for 0 ≤ r ≤ n, with objects all different and no repetition.

Arranging all n distinct objects

nPn = n!

Uses 0! = 1.

Special values

nP0 = 1, nP1 = n

nP1 = n because one place can be filled in n ways.

Always together

(n − k + 1)! × k!

k specified objects out of n, all arranged in a row. Block counts as one object; k! is the inside order.

Not all together (k specified objects)

n! − (n − k + 1)! × k!

Total arrangements minus the arrangements where all k are together. For k = 2 this is the same as 'the two are never together'. For k > 2 it does not mean no two are adjacent.

Two objects never together (shortcut)

(n − 2)! × (n − 1)P2

Arrange the other n − 2 objects, then choose 2 of the n − 1 gaps in order. Gives the same value as n! − 2 × (n − 1)!.

No two of k objects adjacent (gap method)

(n − k)! × (n − k + 1)Pk

Arrange the other n − k objects, which creates n − k + 1 gaps. Place the k objects in different gaps, in order. For k = 2 it matches the shortcut above.

Recurrence

nPr = n × (n − 1)P(r − 1)

Useful for quick checks.

How to solve Permutations of Distinct Objects questions

1. 1Decide if order matters. If the question asks to arrange, rank, form words or numbers, it is a permutation.
2. 2Identify n (total distinct objects available) and r (places to fill).
3. 3Check for repetition. If objects can repeat, the nPr formula does not apply; fill each place separately.
4. 4If there is no restriction, compute nPr = n! ÷ (n − r)!, or multiply r descending terms starting from n.
5. 5For 'always together', form one block of the k objects. Arrange the block with the rest, then multiply by k! for the order inside the block.
6. 6For 'two objects never together' (or 'not all together'), find total arrangements and subtract the 'together' count. If the question means no two of k > 2 objects adjacent, use the gap method instead.
7. 7For restrictions on position (first place fixed, digit not zero first), fill the restricted place first, then the rest.
8. 8Match your answer with an option and check that it is a sensible size.

Quickest way: Place-filling and option elimination

Common mistakes in Permutations of Distinct Objects

• Using nCr instead of nPr.

  Students do not check whether order matters.

  Fix: Swap two items mentally. If the outcome changes, use nPr. Remember nPr = nCr × r!.

• Forgetting the inside arrangement of the block in 'always together'.

  Students stop after arranging the block with other objects.

  Fix: Always multiply by k! for the objects inside the block.

• Computing 'never together' as n! − (n − 1)!.

  Students forget to multiply the together count by the inside order.

  Fix: For two objects use n! − 2 × (n − 1)!. For k objects 'not all together', use n! − (n − k + 1)! × k!.

• Using n! − (n − k + 1)! × k! when the question says no two of k objects can be adjacent.

  For k > 2 this formula only removes the cases where all k are in one block. It still counts arrangements where some two are adjacent.

  Fix: Read the condition carefully. For no two adjacent, use the gap method: (n − k)! × (n − k + 1)Pk.

• Writing nPr = n! ÷ r! or n! ÷ (n − r).

  Mixing up with nCr, or missing the factorial on the denominator.

  Fix: Denominator is (n − r)! only. Test with 5P2 = 20.

• Allowing zero in the first place of a number.

  Students apply nPr directly to digits 0–9.

  Fix: Fill the first place with a non-zero digit first, then fill the remaining places from the leftover digits.

Worked examples