Question #5eb8c

1 Answer
Nov 14, 2017

#(8!)/((8-5)!) =6720#

Explanation:

Since #8# and #5# have already been used,
we need a sequence of #5# digits from the remaining #8# possible digits.

There are
#color(white)("XXX")8# choices for the first digit, leaving
#color(white)("XXX")7# choices for the second digit, leaving
#color(white)("XXX")6# choices for the third digit, leaving
#color(white)("XXX")5# choices for the fourth digit, leaving
#color(white)("XXX")4# choices for the fifth digit.

#8xx7xx6xx5xx4=6720#

This type of question comes up fairly often as the number of permutations of arrangements of #k# items selected from a collection of #m# items:
#color(white)("XXX")P(k,m)=(m!)/((m-k)!)#