Question

How many numbers greater than a million can be formed with the digits \[2,3,0,3,4,2,3\] ?

Answer 1

Hint:In order to solve the given problem, first don't consider the fact of numbers greater than a million. First find out the total number of unique numbers that can be formed by the help of given digits and further find out the number of possible numbers which has zero at the millionth position. Finally to find out the number of required numbers, subtract the number of numbers with zero at millionth position from the total number of possible numbers.

Complete step-by-step answer:
Given digits are:
\[2,3,0,3,4,2,3\]
Now first let us find the total number of possible unique numbers that can be formed from these digits.
We have 7 different digits out of which digit 2 is repeated twice and the digit 3 is repeated thrice amongst the digits.
As we know that number of unique arrangements of p different items out of which x and y number of items are repeated is given by:
\[\dfrac{{p!}}{{x!y!}}\]
Using the above concept for finding the number of unique numbers we get:
Number of unique arrangement is:
\[\dfrac{{7!}}{{2!3!}}\]
Let us simplify the factorial to find the exact value.
\[
   = \dfrac{{7!}}{{2! \times 3!}} \\
   = \dfrac{{7 \times 6 \times 5 \times 4 \times 3!}}{{2! \times 3!}} \\
   = \dfrac{{7 \times 6 \times 5 \times 4}}{2} \\
   = \dfrac{{840}}{2} \\
   = 420 \\
 \]
So we have 420 different arrangements of numbers.
These numbers also have the numbers containing 0 at the millionth position.
We know that the digit at the millionth position is 0. Eventually the number becomes less than a million.
So we need to find the number of such arrangements with 0 at the millionth position.
Let us fix 0 at the millionth position and find the number of such arrangements possible.
Keeping 0 digit at the millionth position we need to arrange 6 different digits out of which 2 is repeated twice and 3 is repeated thrice.
So, number of such arrangements are:
\[\dfrac{{6!}}{{2!3!}}\]
Let us simplify the factorial to find the exact value.
\[
   = \dfrac{{6!}}{{2! \times 3!}} \\
   = \dfrac{{6 \times 5 \times 4 \times 3!}}{{2! \times 3!}} \\
   = \dfrac{{6 \times 5 \times 4}}{2} \\
   = \dfrac{{120}}{2} \\
   = 60 \\
 \]
So, we have 60 numbers with 0 at the millionth position.
Therefore, now let us find out numbers which are greater than a million.
Number of required number = total number of numbers formed by the digits - number of numbers with 0 at million position.
\[
   = 420 - 60 \\
   = 360 \\
 \]
Hence, 360 numbers greater than a million can be formed with the digits given.

Note-:This problem can also be solved by separately considering and fixing all the digits other than 0 at the million position and finding out the number of different arrangements for each of the digits at million positions and finally adding them to get the answer. But that method will be a bit longer than one done above. Students must remember the formula for different types of arrangements with similar items or repeated numbers and also the method for simplifying the factorial.