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3 DIGIT NUMBERS QUESTION

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3 DIGIT NUMBERS QUESTION

by cramya » Sat Sep 27, 2008 9:30 am
Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?

A. 24
B. 36
C. 72
D. 144
E. 216

I am getting 144 but in some of the posts I have seen the answer seems to be 216.

Could some one share their insights please? I am trying to find out what numbers I missed in this count.
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Re: 3 DIGIT NUMBERS QUESTION

by Morgoth » Sat Sep 27, 2008 9:42 am
cramya wrote:Of the three-digit positive integers that have no digits equal
to zero, how many have two digits that are equal to each
other and the remaining digit different from the other two?

A. 24
B. 36
C. 72
D. 144
E. 216

I am getting 144 but in some of the posts I have seen the answer seems to be 216.

Could some one share their insights please? I am trying to find out what numbers I missed in this count.
Well you have to consider three cases here.

2 digits equal and 1 digit different.

Permutations via anagram form

Case I [ABB]

8*9*1 = 72

Case II [BAB]

8*9*1 = 72

Case II [BBA]

8*1*9 = 72

Total integers or permutations = 72+72+72 = 216
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by cramya » Sat Sep 27, 2008 9:47 am
Thanks Morgoth! I am still missing something here.

Whats about the case AAB then?

Eg; There are 9 digits 1-9

I take 12 (the numbers that fit this solution are

112 case AAB
121
212
221

So for each combination of 1 with 2-9 there are 32 numbers.

For 2 there will be 28 (since 1,2 combination numbers are the same as 2,1 combination numbers)

For 3 there will be 24 (3,4 ; 3,5 ; 3,6 ; 3,7 ;3,8 ;3,9) and not taking in to account 3,1 and 3,2 since 1,3 and 2,3 would have included these

On a similar note for 4 there will be 20 etc....
..... Total = 144

I must not be thinking correctly and missing something and thats what I wanted to know.
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by cramya » Sat Sep 27, 2008 9:52 am
Sorry.Read through it again and understood! Thanks :D
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by pre-gmat » Sat Sep 27, 2008 10:04 am
Hi Morgoth,

I am not very clear about your explanation.
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by cramya » Sat Sep 27, 2008 10:15 am
The way I understood is somewhat similar to Morgtoh's explanation but here's how it goes

Fix the digit(1-9) that need to occur twice in the number as B

B can occur in the number as

BBA

The first B can be filled 9 ways(1-9)
The second B 1 way (it has to be the digit that you shoose for first B)
The A can be filled in 8 ways

So 9*1*8 = 72

BAB

9*8*1


ABB
9*8*1

So adding all three 72+72+72 or 72*3 = 216

Hope this help! I was also confused on how to go about this prob initially
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by Morgoth » Sat Sep 27, 2008 10:19 am
cramya wrote:The way I understood is somewhat similar to Morgtoh's explanation but here's how it goes

Fix the digit(1-9) that need to occur twice in the number as B

B can occur in the number as

BBA

The first B can be filled 9 ways(1-9)
The second B 1 way (it has to be the digit that you shoose for first B)
The A can be filled in 8 ways

So 9*1*8 = 72

BAB

9*8*1


ABB
9*8*1

So adding all three 72+72+72 or 72*3 = 216

Hope this help! I was also confused on how to go about this prob initially
Perfect!

This is exactly the way I did.
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by pre-gmat » Sat Sep 27, 2008 10:39 am
perfect...

Thanks....
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by kris610 » Tue Sep 30, 2008 7:24 am
You can select 2 places in 3 possible ways, for each of which you have 9 possible values (No zero). For each of these 9 values you've 8 possible values. 3 * 9 * 8 =216.
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by harchetan » Thu Aug 09, 2012 12:10 pm
I have a different approach of doing this question.

Let us split the total no of ways of making a 3-digit no.(from 1,2,3....9 --> 9 options; ZERO excluded)

Total Ways = (When all digits r same) + (When 2 digits r same) + (when none are same)

=> 9 x 9 x 9 = 9 x 1 x 1 + X + 9 x 8 x 7
=> 729 = 9 + X + 504
=> X = 729 - 504 - 9
=> X = 216 [ANSWER]

Therefore, Correct answer: E[/u]
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