Question 1152140

The first marble can be one of {{{5}}} colors.

For each of these {{{5}}} colors, the second marble can then be one of {{{4 }}}colors.

For each of these {{{5*4}}} combinations, the third marble can be one of{{{ 3 }}}remaining colors.

For each of these {{{5*4*3}}} combinations, the fourth marble can be one of {{{2}}} remaining colors.

For each of these {{{5*4*3*2 }}}combinations, the last marble can only be the {{{1}}} remaining color.

Total combinations: {{{5*4*3*2*1 = 5!  = 120}}}