# Find the number of numbers, greater than a million that can be formed with the digit 2, 3, 0, 3, 4, 2, 3.

Given, the numbers 2, 3, 0, 3, 4, 2, 3. Total of 7 digits, and it has 2 repeated digits 2, and 3 repeating twice and thrice respectively.

To find: Number of seven-digit numbers that can be formed using digits 2, 3, 0, 3, 4, 2, 3. Notice that an arrangement in which digit 0 in the first place will not be counted as seven-digit number, i.e. greater than 1 Million (1000000). For example- 0232334 will not be counted as seven-digit number since it is a 6 digit number.

The problem can now be rephrased as to find total number of permutations of 7 objects (2, 3, 0, 3, 4, 2, 3) of which two objects are of same type (2, 2), and three objects are of another type (3, 3, 3), And all other objects are distinct. But, 0 cannot be in first place (Condition of seven digit number).

Since we know, Permutation of n objects taking r at a time is nPr,and permutation of n objects taking all at a time is n!

And, we also know Permutation of n objects taking all at a time having p objects of same type, q objects of another type, r objects of another type is . i.e. the, number of repeated objects of same type are in denominator multiplication with factorial.

First, we will find a total number of permutations of these 7 digits and then we will go minus all those permutations in which 0 will come in first place. This will give us exactly number of seven-digit numbers that can be formed by permuting the given digits, i.e. 2, 3, 0, 3, 4, 2, 3.

Total number of permutations

Number of permutations in which 0 will come at the first place will be equal to (Number of ways we can arrange the remaining digits, i.e. 2, 3, 3, 4, 2, 3 in the remaining six places)

Total number of permutations of given digits forming a seven-digit number

= 360

Hence, total number of permutations of 7 digits (2, 3, 0, 3, 4, 2, 3) forming a 7 digit number is equals to 360.

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