Question

The number of numbers greater than 400000 that can be formed by using the digit 0, 2, 2, 4, 4, 5 is:
A. 90
B. 32
C. 20
D. None of the above

Answer 1

Hint: This question is based on the concept of permutation of objects which are not all different just like the digits 0, 2, 2, 4, 4, 5 given above. Therefore the permutation of n objects taken all at a time when p objects are alike and of one kind, q objects are alike and of another kind, r objects are alike of yet another kind and the remaining objects are different is
$\dfrac{{\left| \!{\underline {\,
  n \,}} \right. }}{{\left| \!{\underline {\,
  p \,}} \right. \left| \!{\underline {\,
  q \,}} \right. \left| \!{\underline {\,
  r \,}} \right. }}$ .

Complete step-by-step answer:
Since 400000 is a 6-digit number and the number of digits to be used is also six. Also, the numbers have to be greater than 400000, so they can begin with either 4 or 5.
CASE-1 When 4 occupies the extreme left position. Here we have to fill up the remaining 5places with the digits 0, 2,2,4,5 out of which 2-appears twice. This can be done in $\dfrac{{\left| \!{\underline {\,
  5 \,}} \right. }}{{\left| \!{\underline {\,
  2 \,}} \right. }}$ = 60 Ways.
CASE-2: When 5 occupies the extreme left position. Here we have to fill up the remaining 5-places with the digits 0, 2,2,4,4 out of which 2 and 4 appear twice. This can be done in $\dfrac{{\left| \!{\underline {\,
  5 \,}} \right. }}{{\left| \!{\underline {\,
  2 \,}} \right. \left| \!{\underline {\,
  2 \,}} \right. }}$ = 30 Ways.
Hence the required numbers greater than 400000 formed using the above digits $= 60+30 =90$.
So, the correct answer is “Option A”.

Note: We observe that out of the total given digits which are six in number 2 are one of a kind and other 2 are of another kind where the left two digits are both different. Therefore there are a total 4 types of the digits or objects.