How many 5-digit permutations can be formed from the set S={1,2,3,4,5,6,7,8,9} provided the three digits {1, 5, 8} are together in every permutations in any of 3! ways? Thank you.
If we group the 3 digits (1, 5, 8) as one digit, then we have 6 + 1 = 7 digits
1 - If the permutation begins with (1, 5, 8), the we have: 3! x 6 x 5 = 180 permutations
2 - If the permutation begins with one the other 6 digits, then we have: 6 x 3! x 5 =180 permutations
3 - If the permutation begins with 2 of the other 6 digits, then we have: 6 x 5 x 3! =180 permutations =
4 - So, the grand total =180 x 3 =540 permutations.
