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

Kudos for R.A.E

Here,

I will use combination to solve the problem.

From 1982 to 1980 we have = 3 years

SO the number given = 6344409, where first 2 digit is the number of years and then we have remaining 5 digit

so the first 2 digit can be filled up in = 3 ways

The remaining 5 digits can be filled up in \(= 10 * 10 * 10 * 10 * 10\)(since they are not restrictive and we can choose any number from 0 to 9)

SO the total number of ways = \(3 * 10 * 10 * 10 * 10 * 10 = 300,000\)

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