How many 3 digit numbers can we make such that two of the digits are

Question

How many 3 digit numbers can we make such that two of the digits are same and non of the digits equal zero?

A. 60
B. 72
C. 150
D. 216
E. 280

SonalSinha803 · Answer

Following are the possibilities -

**# or *## or *#*
9*8 or 9*8 or 9*8 ways
i.e., 72*3 = 216 ways. Thus, D is the answer.

Posted from my mobile device

LevanKhukhunashvili · Answer

We have to consider several scenarios
1) xxy 9*1*8=72
2) xyx 9*8*1=72
3) yxx 8*9*1=72

3*72=216
Where x is the same digits and y is the different one

In My Opinion
Ans:  D

GyMrAT · Answer

Since the digits are non zero, hence 9 digits {1,2,....9} available to chose from

First digit = 9C1 = 9
Second digit = same as first digit = 1
Third digit = 8C1 = 8

Arrangements possible = 3!/2! = 3

Hence total 3 digit such #'s possible = 9*1*8*3 = 216

Answer D.

Thanks,
GyM

computer-bot · Answer

Thanks everyone for the detailed answer however I have the following few questions.

1. Why use 9C1 and 8C1? Usually when we use Fundamental Counting Principle, we simply put the # of possibilities at each step. If we didn't have the restriction that two digits are the same then we would have solved it by 9*8*7. Is combination generally hidden in FCP?

2. Also why did you say "Arrangements Possible = 3!/2!"? I thought the permutation took care of all arrangements when we compute 9*1*8 (XXY) - as the permutation definition says that  if an event can occur in  m  ways and a second event in  n  ways then the two events can occur in  m*n  ways . When we have the general form XYZ, is 3!/3! hidden in the calculation?

Regards,

GyMrAT · Answer

If you want to master Combinatorics, i suggest, if you have not already, then devote an obscene amount of time in assimilating the concepts explained by Karishma in the below article. Its a Combinatorics Bible.

https://gmatclub.com/forum/combinatoric ... l#p1579442

Hope that helps.

Thanks,
GyM

Blueblu · Answer

Hi mate, U are getting too technical

For a 3 digit number, wherein u cannot have a zero and just 2 numbers repeated, make a slot - _ _ _ In first slot you can have any number from 1 to 9, so you have 9 options there. Second slot (the number being same as slot 1), you have only 1 option. Third slot can have any number other than the one in slot 1 and 2, giving you only 8 options. So you get 9*1*8 = 72 Note, you can have these numbers in 3 orders. Hence multiply 72 by 3. Did that help? Is it any easier for you now?

computer-bot · Answer

Thanks. I haven't gone through it yet but I needed it.

Blueblu... thanks for the response. What I don't understand is that if we had all three distinct digits XYZ, we would have simply multiplied by 9*8*7. But XYZ can also be arranged in the following 6 ways but we don't multiple X*Y*Z with 6.

XYZ
XZY
YXZ
YZX
ZXY 
ZYX

Why is that so????

Blueblu · Answer

You are confusing distinct digits with the number of options. All of your digits X Y and Z are one of the digits from 1 to 9. Say you pick 678 and compare it with 876; in the first slot you put 9 which takes care of all 9 options including the digits 6,7 and 8 and so forth. This is why you don't multiply 9*8*7 with 3. I agree with GymArt - read Karishma's bible to get a clarity of your concepts.

computer-bot · Answer

On it

Thanks

CounterSniper · Answer

step 1 .
two of the digits are same
assume we have 3 places _ _ _
no of ways we can select 2 out of these three to be the same is 3c2 = 3

step 2.
selecting two same digits
no of ways we can select 1 st one = 10 , but it must not be 0 = 9 
no of ways we can select the 2nd one =1

step 3
no of ways we can select the last digit = 8

therefore ,
3*9*1*8=216

BrushMyQuant · Answer

This is a selection problem

Let's select the two values of three which are going to be the same. This can be done in 3C2 ways
=>  \frac{3!}{(2!*1!)}  = 3 ways

Now the digit which is going to be the same can be selected in 9 ways
(we have digits from 1 to 9 to select as digits cannot be 0)

The digit which is different can be select in 8 ways. (all numbers from 1 to 9 except the one selected for the common digits)

So, in total we have 3 * 9 * 8 = 216 ways

So, answer will be D
Hope it helps!

luisdicampo · Answer

Deconstructing the Question

We are forming 3-digit numbers using digits 1 through 9, with no zeros allowed. Exactly two digits must be the same.

Step-by-step

Choose the digit that is repeated:

9

choices.

Choose the different digit:

8

choices.

Arrange the digits:

\frac{3!}{2!} = 3

Total number of numbers:

9 	imes 8 	imes 3 = 216

Answer D

How many 3 digit numbers can we make such that two of the digits are

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

How many 3 digit numbers can we make such that two of the digits are

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards