How many different​ three-digit numbers can be formed using the digits 9, 4, 8, 2, and 6 without​ repetition? For​ example, 448 is not allowed.

Answer 1

try easier problems first

1 digits with 3 digits 3 ways

2 digits with 4 digits 12 ways = 2(4P2) = 2(4x3/2) = 12

ab, ac, ad 3x4 = 12

ba bc bd

ca cb cd

da db dc

3 digits with 5 digits 5x12 = 60 ways = 2(5P3) = 2(5x4x3/2) = 60

12 ways with 1 digit 1st, 5 different digits first 5x12