Question

Part ID codes consist of 4 different letters and 3 different digits. How many codes are possible if the letters must be kept together? For example, 34ABCD9 and WAXT614 are good, but TR67YZ3 is bad.

Answer 1

The number of permutations of letters is:
\[26P4=\frac{26!}{22!}\]
The number of permutations of digits is:
\[10P3=\frac{10!}{7!}\]
The possible positions of each permutation of digits are:
None ahead of the letters
One ahead of the letters
Two ahead of the letters
Three ahead of the letters

Each permutation of letters can be combined with each permutation of digits and each of these combinations can be combined with each of the four possible positions of digits. Therefore the possible number of codes is given by:
\[\frac{26!\times 10!\times 4}{22!\times 7!}\]