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Joined: 21 Sep 2012
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Location: United States
Concentration: Finance, Economics
Schools: CBS '17
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WE:General Management (Consumer Products)
An employee identification code consists of a vowel followed by a 3-di
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10 Dec 2014, 03:24
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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
A) 211
B) 216
C) 1075
D) 1080
E) 2160
An employee identification code consists of a vowel followed by a 3-di
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10 Dec 2014, 05:02
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Expert Reply
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
A) 211
B) 216
C) 1075
D) 1080
E) 2160
For the case when the hundreds digit is repeated: XXY or XYX
X can take 8 values (2, 3, 4, 5, 6, 7, 8, 9).
Y can take 9 values (10 digits minus the one we used for X).
To account for XXY or XYX case we are multiplying by 2.
So, for this case we'd have 8*9*2.
For the case when the tens and units digits are repeated: XYY
X can take 8 values (2, 3, 4, 5, 6, 7, 8, 9).
Y can take 9 values (10 digits minus the one we used for X).
Now, 8*9 also give number 200 and we need numbers greater than 200, so for this case we'd have 8*9 - 1.
Last step, since there are 5 vowels, then the final answer would be 5*(8*9*2 + 8*9 - 1) = 1,075.
Answer: C.
Check Constructing Numbers, Codes and Passwords in our Special Questions Directory.
General Discussion
Re: An employee identification code consists of a vowel followed by a 3-di
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10 Dec 2014, 05:13
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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
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