Question from http://www.sgforums.com/forums/2297/topics/302719
Four letters from the word STATISTICS are taken to form a code-word. How many of these code words contain
i) only S and T
ii) Neither T nor C
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Answer:
i) 3 letters of S and 3 letters of T
SSST type ~ 4!/3! = 4 words
SSTT type ~ 4! / (2! * 2!) = 6 words
STTT type ~ 4!/3! = 4 words
Summing up = 14 words
or you could view it as for each of the letter, you have 2 choices between S or T
which means,
2 * 2 * 2 * 2 words = 16 words
But this includes SSSS and TTTT, so 16 - 2 = 14 words
ii)
Remaining letters: S S S A I I
Num of S
= 3: then either A or I (2 * 4!/3! => 2 choices of either A or I plus number of permutations for the 4 letters)
= 2: then either with AI (4!/2!) or II (4!/{2!*2!})
= 1: then AII (4!/2! permutations for the 4 letters)
Summing up,
we have number of words = 2 * 4!/3! + 4!/2! + 4!/{2!*2!} + 4!/2!
= 8 + 12 + 6 + 12
= 38 words