Question -
Answer -
Given:
The digits 2, 3, 0, 3,4, 2, 3
Total number of digits= 7
We know, zero cannotbe the first digit of the 7 digit numbers.
Number of 6 digitnumber = n!/ (p! × q! × r!) = 6! / (2! 3!) Ways. [2 is repeated twice and 3 isrepeated 3 times]
The total number ofarrangements = 6! / (2! 3!)
= [6×5×4×3×2×1] /(2×3×2)
= 5×4×3×1
= 60
Now, number of 7 digitnumber = n!/ (p! × q! × r!) = 7! / (2! 3!) Ways
The total number ofarrangements = 7! / (2! 3!)
= [7×6×5×4×3×2×1] /(2×3×2)
= 7×5×4×3×1
= 420
So, total numberswhich is greater than 1 million = 420 – 60 = 360
Hence, total number ofarrangements of 7 digits (2, 3, 0, 3, 4, 2, 3) forming a 7 digit number is 360.