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# The numbers that are greater than ${\text{50,000}}$that can be formed by using the digit ${\text{3,5,6,6,7,9}}$isA. 47B. ${\text{48}}$C. $50$D. None of these  Hint: To form the number that are greater than $50,000$ formed by using ${\text{3,5,6,6,7,9}}$. So, here we have to consider the digits that are initiating with ${\text{5,6,7,9}}$. And then add all possible numbers that can be formed using among the selected digits. As a form, the number initiating with ${\text{5,6,7,9}}$and at the remaining four places any random digits can be placed and then form the numbers with all possible cases and add all of them in order to find total formed numbers.

Considering the number which is greater than $50,000$ contains $5$ digits.
we also have to take the $5$ as with digit number because number forms initiating this digits are also greater than ${\text{50,000}}$out of which the digit $6$ is repeated twice in the arrangement of remaining places numbers.
So, the numbers of permutation can be given as $\dfrac{{{\text{5!}}}}{{{\text{2!}}}} = 60$.
But from these arrangements we also have to reject those numbers which begin with $3$ as in that such case, the numbers will be less than $50,000$.
We find all such numbers by fixing $3$ at extreme left space. So remaining $4$ digits can be filled in $\dfrac{{{\text{4!}}}}{{{\text{2!}}}}{\text{ = 12}}$.the division is done because the number six is repeated twice in arrangement.
Hence, required number of ways to obtain the numbers are ${\text{60 - 12 = 48}}$.
Option (b) is our correct answer.

Note: There are possibilities of calculation errors, and as we have to consider numbers greater than 50,000 so we need to consider the initial digit as 5 or greater than 5 from the given digits.
Here we use the concept of permutation. In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements.
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