Solutions to Practice Problems for the Lander Math Contest
Module 9, Solutions (26-28)
26) The order of the chosen people in this problem does not
matter, so we count the number of combinations: 9C4=9!/(5!
27) Two “words” with same letters but different order are
different, that is why we will calculate the number of
permutations (when order matters): 7P5=7!/2!=2520
28) Let us divide the number of permutations by the number
nPk÷nCk = n!/(n-k)!÷n!/[(n-k)!k!]= n!/(n-k)!× [(n-k)!k!]/n!=k!