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A 3-digit positive integer consists of non zero digits. If

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A 3-digit positive integer consists of non zero digits. If  [#permalink]

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New post 15 Aug 2011, 01:22
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A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?
(A) 72
(B) 144
(C) 216
(D) 283
(E) 300
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Re: Probability - Exactly Question!  [#permalink]

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New post 15 Aug 2011, 01:51
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DeeptiM wrote:
A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?
(A) 72
(B) 144
(C) 216
(D) 283
(E) 300





here is the answer in pure probability terms.
let the 3 digit number be XYZ

now none of them contain zero --> they can vary from 1 - 9

so lets take the value of X, X can be from 1 - 9, hence X can be selected in 9 ways
since Y =X, after selecting X you can select Y in 1 way only
Z can be an selected from 1 -9 (minus 1 because it cannot be the same as X) in exactly 8 ways

hence total number of ways of selecting the number is 9 * 8 = 72.

BUT we are asked for all possible numbers that can be formed.

So XYZ ( or XXZ in this case) can be moved around in 3 ways
XXZ
XZX
ZXX

hence the total number of numbers is 72 *3 = 216
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Re: Probability - Exactly Question!  [#permalink]

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New post 15 Aug 2011, 02:00
Ques says..non zero digits..y didnt u consider the negative values?
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Re: Probability - Exactly Question!  [#permalink]

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New post 15 Aug 2011, 02:04
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the first few words in ur ques says "A 3-digit positive integer " :-)

although im not Einstein i dont think we can have a postive 3 digit number with negative digits :-P
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Re: Probability - Exactly Question!  [#permalink]

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New post 15 Aug 2011, 02:07
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Gosh, I cant stop laughing on my sillyness...lol
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Re: Probability - Exactly Question!  [#permalink]

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New post 15 Aug 2011, 02:19
:-) we just have to make sure we dont miss such things on the real test... or we just butchered our test and you might watch your months of preps flowing down the drain!!
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Re: Probability - Exactly Question!  [#permalink]

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New post 15 Aug 2011, 02:23
1
i knwww and the worst part is I keep doing these mistakes..boo hoo..

thanks for all the help Vikas..
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Re: Probability - Exactly Question!  [#permalink]

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New post 15 Aug 2011, 02:28
DeeptiM wrote:
A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?
(A) 72
(B) 144
(C) 216
(D) 283
(E) 300


Same as viks4gmat:
\(C^{9}_{2}*2*\frac{3!}{2!}=36*2*3=216\)

\(C^9_2:\) Number of ways we can select 2 digits out of 9
{1,2,3,4,5,6,7,8,9}
We can select
1,2
OR
4,9

But, we need to multiply each of these selections by 2. Why?

For 8,9
8,8,9
8,9,9
Are two possibilities. First, in which we choose 2 8's and second, in which we choose 2 9's.

If we have;
XXY

It can be uniquely arranged in = 3!/2! ways
3 terms: X,X,Y=3!(numerator)
2 repeats: X,X=2!(denominator)

So, if we select
889
---
889
898
988
Total=3 ways

Ans: "C"
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Re: Probability - Exactly Question!  [#permalink]

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New post 15 Aug 2011, 02:33
DeeptiM wrote:
Ques says..non zero digits..y didnt u consider the negative values?


If GMAT uses the word "digit", it is always non-negative, single digit integers.

There are only 10 possible digits:
{0,1,2,3,4,5,6,7,8,9}

So, non-zero digits
{1,2,3,4,5,6,7,8,9}

A real number -integers, rational/irrational numbers, decimals, etc.- can be negative, positive or zero.
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Re: A 3-digit positive integer consists of non zero digits. If  [#permalink]

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New post 23 Dec 2015, 04:52
DeeptiM wrote:
A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?
(A) 72
(B) 144
(C) 216
(D) 283
(E) 300


i took a different approach but missed something

total possibilities : 9x9x9 = 729
no two digits are the same : 9x8x7 = 504

exactly two = 225 ?

what did i miss?
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Re: A 3-digit positive integer consists of non zero digits. If  [#permalink]

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New post 23 Dec 2015, 05:01
jimwild wrote:
DeeptiM wrote:
A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?
(A) 72
(B) 144
(C) 216
(D) 283
(E) 300


i took a different approach but missed something

total possibilities : 9x9x9 = 729
no two digits are the same : 9x8x7 = 504

exactly two = 225 ?

what did i miss?


your answer 225 consists of all three digit same ..
iy will be 9*1*1=9..
so actual answer=225-9=216..
hope it helps
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Re: A 3-digit positive integer consists of non zero digits. If  [#permalink]

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New post 26 May 2017, 22:10
Top Contributor
DeeptiM wrote:
A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?
(A) 72
(B) 144
(C) 216
(D) 283
(E) 300


1. When units and tens digits are the same. There are 9 such cases. For each, hundreds digit can have 8 values for a total of 72
2. Similarly, units and hundreds and also tens and hundreds can have the same values, for a total of 72*3=216
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Re: A 3-digit positive integer consists of non zero digits. If  [#permalink]

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New post 23 Oct 2018, 04:55
DeeptiM wrote:
A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?
(A) 72
(B) 144
(C) 216
(D) 283
(E) 300


I thought of the question this way..

I first thought of making 1 as the number that appears twice.

There are three ways :- 11x, 1x1 and x11 where x is a number from 2 to 9. We find that there 24 possible ways= 3 ways * 8 (because any number from 2 to 9 can be there in the space of x) = 24.

Now multiply 24 into 9 because now any number from 2 to 9 can be the two common digits.

24 * 9 = 216.
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Re: A 3-digit positive integer consists of non zero digits. If  [#permalink]

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New post 23 Oct 2018, 05:08
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1
DeeptiM wrote:
A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?
(A) 72
(B) 144
(C) 216
(D) 283
(E) 300


We need 3 digit integer with non zero digits (so only 1 to 9 are allowed). Since 2 of the digits are the same, the numbers are of the form XXY or XYX or YXX (so 3 forms).

You can select X in 9 ways and Y in 8 ways.

Total integers possible = 9*8*3 = 216

Answer (C)
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Re: A 3-digit positive integer consists of non zero digits. If  [#permalink]

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New post 16 May 2019, 19:59
DeeptiM wrote:
A 3-digit positive integer consists of non zero digits. If exactly two of the digits are the same, how many such integers are possible?
(A) 72
(B) 144
(C) 216
(D) 283
(E) 300


The 3 digits are nonzero, and they consist of two distinct digits. The number of ways to choose 2 digits from 9 nonzero digits is 9C2 = (9 x 8)/2 = 36. Now for each pair of digits chosen, for example, 1 and 2, we could have: 112, 121, 211, 221, 212, and 122. Therefore, there are 36 x 6 = 216 such integers.

Answer: C
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Re: A 3-digit positive integer consists of non zero digits. If   [#permalink] 16 May 2019, 19:59
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