desaichinmay22 wrote:

An employee identification code consists of a vowel followed by a 3-digit number greater than 200. Exactly 2 of the 3 digits in the code should be identical. How many different codes is it possible to form?

A) 211

B) 216

C) 1075

D) 1080

E) 2160

identification code will be of the type -,-,-,-

first digit can be selected from any of the 5 vowels in 5C1 ways

now for the remaining three digit lets consider the following two cases

case 1: when the number is greater than 200 but less than 300

number will be of the type 2,_,_. now suppose repeating number is same as first digit number i.e. 2. and the third number is one of the remaining 9 numbers (we are rejecting 2 here, because it will result in 222, which is not acceptable as per the given condition). thus these two number can arrange themselves in two blank spaces in 2! ways. hence total number of numbers in which repeating digit is same as the first digit = 1.9.2! =18

now, suppose that repeating number is different than first digit. thus possible case in this case are 8 as listed below:

211

233

244

255

266

277

288

299

here again we have rejected 200( because number must be greater than 200) and 222 ( exactly two repeating digits are allowed)

thus total possible cases are 18 + 8 =26

case 2: number ranging from 300 to 999

here for first digit we have 7 cases (3,4,5,6,7,8,9)

now if the repeating number is same as the first number then we will have 18 cases ( same reasoning as mentioned in the previous case)

if the repeating number is different than first digit number then we will have 9 cases ( because here number ending with two zeros are allowed)

hence total number of ways = 7(18+9) = 189

thus different number of codes = 5(189+26) = 1075

hence C