Permutations and Combinations question

Question

Ten cards, each bearing a single letter, spell out the word PROSPEROUS. 3 cards are selected, and the order of selection is not relevant. Find how many possible selections can be made i) if 2 of the 3 cards bear the same letter ii) without restriction

Answer 1

Count the frequency of different letters: P (2); R (2); O (2); S (2); E (1); U(1). There are 10C3=10*9*8/6=120 ways of picking 3 cards out of 10 if they are all different, and there are 6C3=6*5*4/6=20 ways of picking 3 out of 6 different cards (only 6 different letters are used). But there are 4 letters repeated just once. So the repeated letter can be combined with each of the remaining 5 letters, giving us 4*5=20 further combinations, making 40 in all. Alphabetically, these are (E, O, P, R, S, U):

EOO, EOP, EOR, EOS, EOU, EPP, EPR, EPS, EPU, ERR, ERS, ERU, ESS, ESU, OOP, OOR, OOS, OOU, OPP, OPR, OPS, OPU, ORR, ORS, ORU, OSS, OSU, PPR, PPS, PPU, PRR, PRS, PRU, PSS, PSU, RRS, RRU, RSS, RSU, SSU.

Out of these 40, here are the 4 sets of 5 (20) containing repeated letters (shown in bold above):

EOO, OOP, OOR, OOS, OOU

EPP, OPP, PPR, PPS, PPU

ERR, ORR, PRR, RRS, RRU

ESS, OSS, PSS, RSS, SSU

So the answer to i) is 20 and ii) 40.