How many positive integers 'n' can be form using the digits 3, 4, 4, 5, 6, 6, 7 if we want 'n' to exceed 60,00,000?
A. 320
B. 360
C. 540
D. 720
Answer: Option C
Solution(By Examveda Team)
As per the given condition, number in the highest position should be either 6 or 7, which can be done in 2 ways. If the first digit is 6, the other digits can be arranged in $$\frac{{6!}}{{2!}}$$ = 360 ways. If the first digit is 7, the other digits can be arranged in $$\frac{{6!}}{{2! \times 2!}}$$ = 180 ways. Thus required possibilities for n, = 360 + 180= 540 ways
This Question Belongs to Arithmetic Ability >> Permutation And Combination
It is apparent that the first digit must be 5 , 6 or 7 . There are a total of 7 digits out of which we have a total of 4 of 5s , 6s and 7s
Hence, 47 of the total number of positive integers that can be formed from the given digits obey the constraint.
The total number of positive integers that can be formed from the given 7 digits noting that 4 and 5 appear twice
=7!2!2!
∴ The total number of positive integers satisfying the constraint
=47×7!2!2!
=47×7×6!4
=6!
=720