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An employee identification code consists of a vowel followed by a 3-di

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An employee identification code consists of a vowel followed by a 3-di [#permalink]

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New post 10 Dec 2014, 02:24
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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
[Reveal] Spoiler: OA

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An employee identification code consists of a vowel followed by a 3-di [#permalink]

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New post 10 Dec 2014, 04:02
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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


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.

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Re: An employee identification code consists of a vowel followed by a 3-di [#permalink]

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New post 10 Dec 2014, 04:13
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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


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

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Re: An employee identification code consists of a vowel followed by a 3-di [#permalink]

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New post 10 Dec 2014, 07:41
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.

--------------------------------------------------------------------------------------------------------------------------------------------------
Hi Bunuel

Can you plz explain the highlighted part why did we minus 1 and how does it gives number 200.


Regards
SG

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An employee identification code consists of a vowel followed by a 3-di [#permalink]

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New post 10 Dec 2014, 09:26
Can you plz explain the highlighted part why did we minus 1 and how does it gives number 200.


Regards
SG[/quote]


As X can take any number from (2, 3, 4, 5, 6, 7, 8, 9) and Y can take any number other than the one chosen for X.

So 200(X=2,Y=0,Y=0) is also included in 8*9 , hence we have to subtract 1

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Re: An employee identification code consists of a vowel followed by a 3-di [#permalink]

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New post 10 Apr 2015, 12:22
Very elegant solution Bunuel. An alternate solution:

Required no. of ways (R)=Total no. of ways - (no. ways of ways when all digits are different + no. of ways when 3 digits are identical) -5 cases when vowel is combined with 200
R=5*8*10*10 - (5*8*9*8 + 5*8) - 5
= 5*8(100-72-1)-5
=40*27-5=1080-5=1075

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Re: An employee identification code consists of a vowel followed by a 3-di [#permalink]

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New post 10 Apr 2015, 21:11
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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


total number of 3-digit number greater than 200 are 799( 8*10*10 -1 )
numbers of the form XXX (all 3 repeated, 222, 333 ...) = 8
numbers of the form XYZ (all 3 digits different) =8*9*8 = 576
so numbers which will have 2 digit repeated are 799 - 8 - 576 = 215

finally considering 5 vowels we will have 5*215 = 1075 as answer
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Re: An employee identification code consists of a vowel followed by a 3-di [#permalink]

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Re: An employee identification code consists of a vowel followed by a 3-di   [#permalink] 12 Oct 2017, 10:18
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