How many permutations of the letters A to H if the three letters ABC must appear together but not necessarily consecutively?

Question

How many permutations of the letters A to H exist if the three letters ABC must appear together but not necessarily consecutively?

Pretty confused on how to start... Can I group ABC as one letter? Then there will be 6 "letters", making it 6!, and then multiplied by 3! because there are 3! ways in which ABC can be assigned.

Answer 1

Yes, that's a good strategy. The trickiest part of combinatorics is figuring out to describe a complicated situation in ways that are relatively easy to count. That's what you did here -- you want to arrange the "letters" ABC, D, E, F, G, H into words, which can be done in 6! ways, and then you can arrange the block ABC in 3! ways for a total of 6!3! = 4320 arrangements.

Answer 2

$A,B,C,D,E,F,G,H$

$ABC$ always appear together. Consider this as a single letter X.

$X,D,E,F,G,H$

Number of permutation=$6!$=$720$

The number of permutation in X is also $3!$.

Therefore total permutations=$6! \times 3!$=$720 \times 6$=$4320$