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Question

The number of integers greater than $$6,000$$ that can be formed, using the digits $$3, 5, 6, 7$$ and $$8$$, without repetition, is


A
216
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B
192
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C
120
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D
72
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Solution

The correct option is B $$192$$
We can have either $$4$$ digit or $$5$$ digit number.

First consider $$4$$ digit number.

We can take $$6,7,8$$ at the thousandth place.

The next three places can be filled in $$^4{C}_3 \cdot 3!$$ way.

So we get $$72$$ such numbers

For $$5$$ digit number, we can choose them in $$5!$$ ways$$(120)$$.

So total no of such numbers are $$72+120=192$$.

Mathematics

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