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How many different 4-letter permutations can be formed from the letters in the word WESTINGHOUSE?
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It is clear that the correct question is to find the number of all distinguishable 4-letter arrangements
(not permutations).
So, I will solve the problem in this formulation.
The given word consists of 12 letters.
Of them, 8 letters are non-repeating; two letters (E and S) are repeating and have multiplicity 2.
So, I first consider 10 unique letters W, E, S, T, I, N, G, H, O, U and will calculate the number of 4-letter
arrangements of these letters. All these arrangements consist of non-repeating letters.
Then I will consider all 4-letter distinguishable words having two repeating "E" and no repeating "S";
then I will consider all 4-letter distinguishable words having two repeating "S" and no repeating "E";
then I will consider all 4-letter distinguishable words having two repeating "E" and two repeating "S".
At the end, I will add all these opportunities.
(1) the number of 4-letter arrangements of 10 distinct letters W, E, S, T, I, N, G, H, O, U is
10*9*8*7 = 5040.
(2) the number of all 4-letter distinguishable words having two repeating "E" is
= = 6*9*8 = 432.
Here = = 6 is the number ow ways to select two positions for E in the 4-letter word,
without looking the order. Next, the factors 9 and 8 are to calculate the number of placing the rest unique letters
at the remaining two positions in the 4-letter word.
(3) the number of all 4-letter distinguishable words having two repeating "S" is
= = 6*9*8 = 432.
Here = = 6 is the number ow ways to select two positions for S in the 4-letter word,
without looking the order. Next, the factors 9 and 8 are to calculate the number of placing the rest unique letters
at the remaining two positions in the 4-letter word.
(4) the number of 4-letter distinguishable words having two repeating "E" and two repeating "S" is
= = 6.
Finally, I sum up the found numbers 5040 + 432 + 432 + 6 = 5910.
ANSWER. There are 5910 such distinguishable 4-letter words.
Solved.