The number of numbers greater than $ 50,000 $ that can be formed by using the digits $ 5,6,6,7,9 $ is
A. $ 36 $
B. $ 48 $
C. \[54\]
D. $ 60 $

Hint: Number of permutations of ‘n’ things taken all together when the things are not all different
To find the number of permutations of things taken all at a time when of them are similar and are of second type of them are similar and are of third type and rest are all different.
So, No. of permutations $ = \dfrac{{n!}}{{p!q!r!}} $
Fractional rotation:
The product of first ‘n’ natural number is denoted by $ n! $ and is read as “dfractional
Thus, \[n! = 1 \times 2 \times 3 \times 4 \times ...... \times (n - 1) \times n\]
Ex. $ 5! = 1 \times 2 \times 3 \times 4 \times 5 $
 $ = 120 $

Complete step-by-step answer:
Permutations and combinations, the various ways in which objects from a set may be selected, generally without replacement, to form subsets. ... This selection of subsets is called a permutation when the order of selection is a factor, a combination when order is not a factor.
Here, we have to find number of numbers greater than $ 50000 $
So, the first digit must be greater than or equal to 5.
Given digits: $ 5,6,7,9 $

So, permutation of $ 5 $ digits i.e. $ 5,6,6,7,9 $ taken all at time where, 2 digits i.e. (6,6) are same is given by $ = \dfrac{{5!}}{{2!}} = \dfrac{{5 \times 4 \times 3 \times 2 \times 1}}{2} = 60 $
 $ \Rightarrow $ total numbers formed $ = \dfrac{{120}}{2} = 60 $
Here, total $ 60 $ numbers can be formed using digits $ 5,6,7,9 $ and all $ 60 $ are greater than $ 50000 $ .
So, the correct answer is “Option D”.

Note: If $ 'n' $ things are different, then the number of permutations is $ n! $
 $ n! = 1 \times 2 \times 3 \times 4 \times 5.......(n - 1) \times n $
 $ ^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}} $
 $ ^n{\operatorname{P} _r} = \dfrac{{n!}}{{(n - r)!}} $
We just take permutations of 5!/2! because one can notice that no matter how you arrange any number formed will be greater than 50,000 even if we find the least number it will be greater than 50,000.