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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?
Answer & Solution
Correct Answer:
Option
C
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
= 540 ways
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LoginHence, 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