Let the number be ABC

We MUST have 2 and 6 in this number. So first let's see how many ways we can do that.

Split the problem into 2 parts:
1. How many ways can we arrange these two digits 2 and 6 which must compulsorily be there, into 3 different places, where position/order matters?
2. How many ways can we choose a third digit?

We have 3 places in which we must ARRANGE 2 digits. So this would be done in 3P2 ways = 6

There is now a third place, which can (in all 6 of those above-mentioned cases) take 9 values 1-9

So in effect we have 6*9 = 54 different three-digit numbers containing both 2 and 6.

Pick B.

albatross86 wrote:Let the number be ABC

We MUST have 2 and 6 in this number. So first let's see how many ways we can do that.

Split the problem into 2 parts:
1. How many ways can we arrange these two digits 2 and 6 which must compulsorily be there, into 3 different places, where position/order matters?
2. How many ways can we choose a third digit?

We have 3 places in which we must ARRANGE 2 digits. So this would be done in 3P2 ways = 6

There is now a third place, which can (in all 6 of those above-mentioned cases) take 9 values 1-9

So in effect we have 6*9 = 54 different three-digit numbers containing both 2 and 6.

Pick B.

That's what I thought. But the OA is A.

The explanation is very vague in the document and it just states that there are 14 possibilities for nos. in their 100s, 300s, 400s, 500s, 700s, 800s, and 900s ( 2 nos. each - 126 and 162 for 100s 326 and 362 for 300s....and so on) - So 14 possibilities.
For 200s, there are 19 possibilities and similarly for 600s, there are 19 possibilities.
So the total no. of outcomes are 14+19+19=52. There is no elaborate explanation on this.

Let me know what you make of that cause i'm totally confused.

Looks like I'm rusty!

I guess we are overcounting by this method in some way though I'm finding it difficult to put my finger on.
I'm guessing I screwed up in splitting the problem up somehow.

There is a detailed explanation here: https://www.gmatdaily.com/20091030-answer.html

Let me know if that is sufficient.

albatross86 wrote:Looks like I'm rusty!

I guess we are overcounting by this method in some way though I'm finding it difficult to put my finger on.
I'm guessing I screwed up in splitting the problem up somehow.

There is a detailed explanation here: https://www.gmatdaily.com/20091030-answer.html

Let me know if that is sufficient.

This is the similar to the explanation I have.....unable to understand it.... shouldn't there be a better approach....I doubt I'll have enough time in the GMAT to solve questions like these in the manner of the explanation.

Hi,
Whats wrong doing it this way ?

_ _ _
now 2, 6, x where x can be any number can be arranged in 6 ways.
Now there are 10 values of x (0-9).
Therefore 10*6=60 ways.
But this includes the possibility of x being zero i.e
X 2 6 and x as zero but we want only 3 digit numbers.

026 and 062 are the only such numbers.
Therefore 60-2=58.

I don't know what I am missing...

singhsa wrote:82. How many different three-digit numbers contain both the digit 2 and the digit 6?
(A) 52
(B) 54
(C) 56
(D) 60
(E) 62

OA later

Case 1: 2 digits are the same.
226,262,622,662,626,266 = 6 options.

Case 2: All 3 digits are different.
Number of options for the third digit = 8. (Any digit other than 2 or 6.)
Number of ways to arrange the 3 digits = 3! = 6.
To combine the options above, we multiply:
8*6 = 48.
Bad integers are 026 and 062.
Subtracting the 2 bad integers:
48-2 = 46 options.

Total options = 6+46 = 52.

The correct answer is A.

TOPGMAT wrote:Hi,
Whats wrong doing it this way ?

_ _ _
now 2, 6, x where x can be any number can be arranged in 6 ways.
Now there are 10 values of x (0-9).
Therefore 10*6=60 ways.
But this includes the possibility of x being zero i.e
X 2 6 and x as zero but we want only 3 digit numbers.

026 and 062 are the only such numbers.
Therefore 60-2=58.

I don't know what I am missing...

The approach above double-counts the integers that include two 2's or two 6's: 226,262,622,662,626,266.

For example, 226 is included when the following are counted:
x26 (where x = the hundreds digit 2)
2x6 (where x = the tens digit 2).

Since the 6 integers above have been double-counted, we must subtract 6 from the total:
58-6 = 52.

3P1=6 ways to arrange two digits in the three-digit number
6N2 -> 1*10*1 (a digit permutation in three-digit number)
2N6 -> 1*10*1
N26 -> 9*1*1
N62 -> 9*1*1
26N -> 1*1*10
62N -> 1*1*10

total 10+10+9+9+10+10=58 less 6 repeated numbers within 6 arrangements (from left to the right, that's why not 12 but 6 only)= 52

singhsa wrote:82. How many different three-digit numbers contain both the digit 2 and the digit 6?
(A) 52
(B) 54
(C) 56
(D) 60
(E) 62

OA later

we have 2, 6 and the rest is any of 8 digits

1.assume that our first number is 2 ,the 2nd one is 6, then the 3d one is 8 (10-2) total=1*1*8=8
2.assume that our 1st number is 6, the 2nd-2, so the rest is any of 8 numbers total=1*1*8
3.assume that our 1st number is 9 (1-9, 0 is excluded) then the 2nd is 2, the 3d is 6 total=9*1*1
4.assume that our 1st number is 9 (1-9, 0 is excluded) then the 2nd is 6, the 3d is 2 total=9*1*1
5.assume that our 1st number is 2,the 2nd is any of 10 numbers, then the 3d one is 6 total=1*10*1
6.assume that our 1st number is 6,the 2nd is any of 10 numbers, then the 3d one is 2 total=1*10*1

so, we have 8+8+9+9+10+10=54

now lets pay attention to the fact, that in (3) and (5) cases we have 226 in common, and in (4) and (6) cases we have 622 in common. so subtracting these 2 numbers from 54 we get 52

My approach:

Three digits _ _ _

Hundreds place .. can be anything from 1-9 => 9 ways
Tens place .. can be anything between 2 and 6 => 2 ways
Ones place .. has to be the number which was not selected in tens place => 1 way

so total no. of ways is 9x2x1 = 18

Now three digits can be arranged themselves in 3! ways.

So the number of digits possible is 18 x 3! => 18 x 6 = 108 ways.

What am I missing? I think I am counting a number multiple times. How do I approach this in right way?

albatross86 wrote:Let the number be ABC

We MUST have 2 and 6 in this number. So first let's see how many ways we can do that.

Split the problem into 2 parts:
1. How many ways can we arrange these two digits 2 and 6 which must compulsorily be there, into 3 different places, where position/order matters?
2. How many ways can we choose a third digit?

We have 3 places in which we must ARRANGE 2 digits. So this would be done in 3P2 ways = 6

There is now a third place, which can (in all 6 of those above-mentioned cases) take 9 values 1-9

So in effect we have 6*9 = 54 different three-digit numbers containing both 2 and 6.

Pick B.

Albatross, I've been seeing this quite a bit and was curious what type of math this is classified under?

"3P2"

First, what is the abbreviation stand for, and how can I learn more about setting up these types of problems i.e. if this were to be found in the manhattan gmat, which book would it be classified in? thanks in advance!

GMATGuruNY wrote:

TOPGMAT wrote:Hi,
Whats wrong doing it this way ?

_ _ _
now 2, 6, x where x can be any number can be arranged in 6 ways.
Now there are 10 values of x (0-9).
Therefore 10*6=60 ways.
But this includes the possibility of x being zero i.e
X 2 6 and x as zero but we want only 3 digit numbers.

026 and 062 are the only such numbers.
Therefore 60-2=58.

I don't know what I am missing...

The approach above double-counts the integers that include two 2's or two 6's: 226,262,622,662,626,266.

For example, 226 is included when the following are counted:
x26 (where x = the hundreds digit 2)
2x6 (where x = the tens digit 2).

Since the 6 integers above have been double-counted, we must subtract 6 from the total:
58-6 = 52.

Thanks a lot gmatguru!!!