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Bunuel
In a puzzle game, Tran was given the word STRUM. How many unique three-letter arrangements are possible using the letters of the word?

A. 15

B. 60

C. 90

D. 120

E. 240

why repetitions are not allowed. It can be 5*5*5 = 125 also.

IIMC ; see highlighted part
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Number of unique three-letter arrangements are possible using the letters of the STRUM
= 5C3*3!
=10*6=60

Bunuel
In a puzzle game, Tran was given the word STRUM. How many unique three-letter arrangements are possible using the letters of the word?

A. 15

B. 60

C. 90

D. 120

E. 240
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Bunuel
In a puzzle game, Tran was given the word STRUM. How many unique three-letter arrangements are possible using the letters of the word?

A. 15

B. 60

C. 90

D. 120

E. 240

why repetitions are not allowed. It can be 5*5*5 = 125 also.

IIMC ; see highlighted part

THAT DOESNT MEAN IT SAYING NO REPETITION.?
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nick1816
Number of unique three-letter arrangements are possible using the letters of the STRUM
= 5C3*3!
=10*6=60

Bunuel
In a puzzle game, Tran was given the word STRUM. How many unique three-letter arrangements are possible using the letters of the word?

A. 15

B. 60

C. 90

D. 120

E. 240


Hi Can you please tell me why did we multiply after 5C3 with 3! ?? and why is not 5P3?
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We can directly find the permutations by using 5P3=\(\frac{5!}{(5-3)!}\)=60

Or
We can choose 3 distinct letters from STRUM. Number of ways= 5C3. Then, we arrange those 3 letters. Total possible arrangements of 3 letters= 3!
Total unique three-letter arrangements possible= 5C3*3!=60

ankitprad
nick1816
Number of unique three-letter arrangements are possible using the letters of the STRUM
= 5C3*3!
=10*6=60

Bunuel
In a puzzle game, Tran was given the word STRUM. How many unique three-letter arrangements are possible using the letters of the word?

A. 15

B. 60

C. 90

D. 120

E. 240


Hi Can you please tell me why did we multiply after 5C3 with 3! ?? and why is not 5P3?
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