Question

Compute the permutation.

How many positive integers of 3 digits each can be formed with the digits 1, 8, 9, 2, 7, 6, 4, and 3, if no digit is repeated in a number?

Answers:
27,
336,
512

Answer 1

So you have 8 digits (number available) to form 3 digit numbers (number selected).

Using the formula $^{a} P_{c}$ where a is the number of available digits, P is the permutation function and c is the number of digits to be selected.

The number of three digit numbers that can be formed = $^{8} P_{3}$
                                                                                       = 336

                                                 O R

By using the formula $Permutation = \frac{a ! }{( a - c) !}$ where a is the number of available digits and c is the number of digits to be selected.

⇒  $Permutation = \frac{8 ! }{( 8 - 3) !}$

⇒  $Permutation = \frac{40, 320 }{120}$

⇒ Permutation = 336

Answer 2

Answer:

336

Step-by-step explanation: