VJesus12 wrote:A code is to be made by arranging 7 letters. Three of the letters used will be the letter A, two of the letters used will be the letter B, one of the letters used will be the letter C, and one of the letters used will be the letter D. If there is only one way to present each letter, how many different codes are possible?
A. 42
B. 210
C. 420
D. 840
E. 5,040
The clause in red seems to imply that repeated letters are IDENTICAL.
Thus, the three A's are identical and the two B's are identical
Number of ways to arrange 7 letters = 7!.
But when an arrangement includes IDENTICAL elements, we must divide by the number of ways each set of identical elements can be ARRANGED.
The reason:
When the identical elements swap positions, the arrangement doesn't change.
Here, we must divide by 3! to account for the three identical A's and by 2! to account for the two identical B's:
7!/(3!2!) = 420.
The correct answer is
C.
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