Question

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

Answer 1

Hint: To solve this question we will use permutation which can be used for counting the number of arrangements possible. Also zero cannot be placed at the first place because if we place the zero at first place the number will become a six digit number.

Complete step-by-step answer:

Here one million $ = 1,000,000.$
Given numbers are \[2,3,0,3,4,2,3\]
A number greater than $1,000,000$ has 7 places, and all the digits are to be used. But 2 is repeated twice and 3 is repeated thrice.
Total number of ways arranging 7 digits
$
   = \dfrac{{7!}}{{(2!3!)}} = \dfrac{{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 3 \times 2 \times 1}} \\
   = 7 \times 6 \times 5 \times 2 \\
   = 420 \\
 $
But the numbers beginning with zero are no more seven-digit numbers then reject those numbers.
Therefore, these can be arranged in
$
   = \dfrac{{6!}}{{(2!3!)}} \\
   = 60 \\
 $
Therefore, the required number of arrangements $ = 420 - 60 = 360$
Hence, 360 numbers can be formed with the given number.

Note: Permutation is an arrangement of sequence or linear order, or if set is already ordered, a rearrangement of its element. The number of permutations of n distinct objects is n factorial, which means the product of positive integers less than or equal to n. Also remember that a repetition of numbers or objects is taken care of by dividing the permutation by the factorial of the number of objects that are identical.