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

Question

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

KarishmaB · Accepted Answer

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)

viks4gmat · Answer

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

DeeptiM · Answer

Ques says..non zero digits..y didnt u consider the negative values?

viks4gmat · Answer

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

DeeptiM · Answer

Gosh, I cant stop laughing on my sillyness...lol

viks4gmat · Answer

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!!

DeeptiM · Answer

i knwww and the worst part is I keep doing these mistakes..boo hoo..

thanks for all the help Vikas..

fluke · Answer

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"

fluke · Answer

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.

jimwild · Answer

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?

chetan2u · Answer

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

SVaidyaraman · Answer

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

hongg7 · Answer

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.

ScottTargetTestPrep · Answer

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

fireagablast · Answer

Three digit number that has two repeats, 1-9 allowed
Total selectable numbers = 9

Choose a number that will repeat: 9C1
Choose a second number: 8 choices left
Arrange them over the spaces: 3!
Divide by the number of repeats: 2!

9C1*8*3!/2!
9*8*3 = 216

Kritisood · Answer

total possibilities = 9*8*7 = 729

in a three-digit number there are only three possibilities:
1. all three digits are the same 9*1*1 = 9 possibilities
2. all three digits are diff 9*8*7= 505 possibilities
3. two digits are the same and one is diff

add 1 and 2 = 513
total - 513 =  216

Sabby · Answer

Hi,

First of all, the way of selecting a 3-digit numbers with 2 numbers that are the same is 3!/2! = 3ways

The way of selecting a 3 digit number with non zero numbers is 9*8*1 in the case of ABA, which gives us 72 ways*3ways=216

Answer C)

Crytiocanalyst · Answer

The key to solving the question is to get the right combination

in order to get the right answer we need to figure out the total no of non-zero numbers that's

9*8*3 since first number we can chose in 9 different ways

one number has to be same hence 1

and chosing the other number in 8 other ways since its different from the given number

3 since they can be in the first middle or last respectively

Therefore IMO 216

summerbummer · Answer

-------------------------------------
 Hi, why's the following approach incorrect?

We know _ _ _ <- is a non-0; 2 digits are same
 ___
Now, we consider 2 blanks as one _ |_ _|

So this can be done in 9 x 8 x 2 ways = 144.

Please correct me!

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

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