Question -
Answer -
(i) 8P3
We know that, 8P3 canbe written as P (8, 3)
By using the formula,
P (n, r) = n!/(n – r)!
P (8, 3) = 8!/(8 – 3)!
= 8!/5!
= (8 × 7 × 6 × 5!)/5!
= 8 × 7 × 6
= 336
∴ 8P3 =336
(ii) 10P4
We know that, 10P4 canbe written as P (10, 4)
By using the formula,
P (n, r) = n!/(n – r)!
P (10, 4) = 10!/(10 –4)!
= 10!/6!
= (10 × 9 × 8 × 7 ×6!)/6!
= 10 × 9 × 8 × 7
= 5040
∴ 10P4 =5040
(iii) 6P6
We know that, 6P6 canbe written as P (6, 6)
By using the formula,
P (n, r) = n!/(n – r)!
P (6, 6) = 6!/(6 – 6)!
= 6!/0!
= (6 × 5 × 4 × 3 × 2 ×1)/1 [Since, 0! = 1]
= 6 × 5 × 4 × 3 × 2 ×1
= 720
∴ 6P6 =720
(iv) P (6, 4)
By using the formula,
P (n, r) = n!/(n – r)!
P (6, 4) = 6!/(6 – 4)!
= 6!/2!
= (6 × 5 × 4 × 3 ×2!)/2!
= 6 × 5 × 4 × 3
= 360
∴ P (6, 4) = 360