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A certain code contains five letters, including one A, two B, and 2 C.

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Re: A certain code contains five letters, including one A, two B, and 2 C. [#permalink]

The total number of arrangements of one A, 2 B's and 2 C's = 5!

As B and C are occurring twice, hence number of ways to create the code = $$\frac{5!}{2!*2!}$$
=> 30
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Re: A certain code contains five letters, including one A, two B, and 2 C. [#permalink]
Bunuel wrote:
A certain code contains five letters, including one A, two B, and 2 C. In how many possible ways can we create the codes?

(A) 5
(B) 10
(C) 12
(D) 30
(E) 60

Solution:

We use the indistinguishable permutations formula, which in this case takes into account the two identical Bs and the two identical Cs. Thus, the number of codes that can be created is 5! / (2! x 2!) = 120/4 = 30.

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Re: A certain code contains five letters, including one A, two B, and 2 C. [#permalink]
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Re: A certain code contains five letters, including one A, two B, and 2 C. [#permalink]
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