carcass wrote:

In a certain nation, every citizen is assigned an identification number consisting of the last two digits of the person’s birth year, followed by five other numerical digits. For instance, a person born in 1963 could have the identification number 6344409. How many identification numbers are possible for people born in the years 1980–1982, inclusive?

(A) 360

(B) 2,880

(C) 288,800

(D) 300,000

(E) 2,400,000

The Answer is D1980-82 that means 3 years: 1980, 1981, 1982.

Also the number of such identification numbres for each year will remain the same. Thus we will calculate the number for 1 year and then multiply it by 3.

For the year 1980

We have 07 slots: __ __ __ __ __ __ __

Out of which the first 2 are fixed:

8 0 __ __ __ __ __

Now, since the numbers can be repeated, we can fill each of the remaining 05 slots in 10 ways(0-9) each.

Hence the total Number of such IDs for 1980 is: 1x1x10x10x10x10x10 = 100000

Now since we need to do the same for 3 years

we can safely calculate the grand total: 3x100000

=3,00,000

_________________

~R.

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