Beatrice Chua's answer to Cell tyh's Junior College 2 H2 Maths Singapore question.
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I hope I didnt miss out any combinations! The answer should be 13 for part (i) and 150 for part (ii)
Date Posted:
5 years ago
… for arrangement with double A,
there are still 4 different letters remaining
(counting 2 same Ns as 1 letter).
4C2 ways to select 2 letters from 4. therefore,
no. of ways = 4C2 x 4!/2! = 72
… similarly for arrangements with double N,
no. of ways = 4C2 x 4!/2! = 72
total no. of ways = 6 + 72 + 72 + 120 = 270.
… and you did remember to multiply the 5C4 for last case of all letters different . . . . . . hmmmm