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How many different six-digit numbers can be formed using all
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19 Jan 2024, 00:16
2
We have to form 6-digit numbers with six given digits - \(3,3,4,4,4,5\)
So it is \(6P6 = \frac{6!}{(6-6)!} =\frac{6!}{0!} = 6!\) since \(0!\) is defined as \(1\).
But we have two \(3s\) and three \(4s\). Since they are indistinguishable when they appear next to each other, we have to eliminate all those instances where they appear next to each other and count them only once. We can do this by dividing \(6!\) by \(2!\) times for the two \(3s\) and \(3!\) for the three \(5s\).
So we finally have = \(\frac{6!}{2! \times 3!} = 60\)
The answer is Choice E.
gmatclubot
How many different six-digit numbers can be formed using all [#permalink]