How many different 3-digit numbers are greater than 299 and do not con

Question

How many different 3-digit numbers are greater than 299 and do not contain the digits 1, 6, or 8?

(A) 222

(B) 245

(C) 291

(D) 315

(E) 343

GMATinsight · Accepted Answer

3 digits number greater than 299 = 999-299 = 700

Three places to be filled to make desired 3 digit number = - - -

Total ways to fill the left most place = 5 choices {3, 4, 5, 7, 9}
Total ways to fill the Middle place = 7 choices {All digits except 1, 6 and 8}
Total ways to fill the Middle place = 7 choices {All digits except 1, 6 and 8}

Total Desired numbers = 5*7*7 = 245

Answer: Option B

Luckisnoexcuse · Answer

Total numbers from 299 to 1000 (excluding both) = 999-300+1 = 700
hundred digit = 3,4,5,7 or 9 = 5 options
tenth digit = 7 options (0,2,3,4,5,7,9)
unit digit = 7 options (0,2,3,4,5,7,9)
5*7*7 = 245
B

generis · Answer

I didn't calculate total number of three-digit integers between 300 and 1,000.

Using slot method to find three-digit numbers that are > 299 and that do not contain 1, 6, and 8

Integers to use if no other restrictions:

0 ,  1 ,  2 ,  3 ,  4 ,  5 , 6 ,  7 ,  8 , 9 :

7 possibilities  if no other restrictions

____ * ____ * ____

FIRST SLOT: 
1) Numbers must be greater than 299. First is 300. So first slot's smallest number is 3
2) Not allowed: 0 and 2. Strike them from 7 total above, =  5  possibilities: 3, 4, 5, 7, 9

SECOND SLOT: only first restriction applies. Hence  7  possibilities for second slot

THIRD SLOT: same as second slot

__ 5 __ * __ 7 __ * __ 7 __ = 245

Answer B

rocko911 · Answer

I dont get it

what if our first digit (hundenredth digit )is 3
it means now we need (0,2,4,5,7,9) on tenth , WHY DID YOU TOOK 3 for tenth digit? Its MENTIONED WE NEED DIFFERENT 3-digit number which means all digit must be different

Please help

niks18 · Answer

Hi  rocko911

the question says  different three digit number  and not different digits. for eg. 333 is different from 444, however the digits in both these numbers are same.

EMPOWERgmatRichC · Answer

Hi All,

This prompt is essentially just a 'permutation' question, but it does have some minor 'quirks' to it that you have to be careful about to get the correct answer.

We're asked for the total number of 3-digit numbers that: 
1) Can only use the digits 0, 2, 3, 4, 5, 7, 9 
2) Must be GREATER than 299.

These restrictions will obviously limit the number of possibilities, but there's one other restriction that we have to remember: a 3-digit number CANNOT begin with a 0 (e.g. 045 is "45", which is a 2-digit number).

The first digit can be: 3, 4, 5, 7, 9 (5 options) 
The second digit can be: 0, 2, 3, 4, 5, 7, 9 (7 options) 
The third digit can be: 0, 2, 3, 4, 5, 7, 9 (7 options)

(5)(7)(7) = 245 total options

Final Answer:  B

GMAT assassins aren't born, they're made, 
Rich

Sajjad1093 · Answer

I get the above method.

But what i initially tried is :

for every "hundreds" 3,4,5,7,9. I calculated combinations for "tens" and "units" places for each of them and add them all up.

like 3 _ _ is 6C2 which is 30.

30*5=150.

What am i doing wrong?

EMPOWERgmatRichC · Answer

Hi Sajjad1093,

There are a few errors in your approach:

1) There are actually 7 digits that could be placed in the tens and units 'spots' (not 6).
2) The number 305 and 350 are DIFFERENT results that you have to account for, so this is a permutation (NOT a combination). 
3) Digits can be used repeatedly (for example, 333 and 355 are possible options that you have to consider).

GMAT assassins aren't born, they're made,
Rich

Sajjad1093 · Answer

my bad i didnt write clearly.

so i have assumed the following:

1) I considered "different 3-digit numbers" is all digits being different. like 334 or 332 not allowed. 302 and 320 are allowed becuase each of the 3 digit is different. But now you have stated that digits can be repeated, its the figure that needs to be different.

2) So even considering the above correction i have done  7!  because considering the limitation on the digits to be used which is 1,6,8 being off limit which leaves us with 7 digits.

I have also applied permutations and not combinations. As i have calculated 370 and 307 as unique arrangement

Like for 3 _ _ i have calculated  7!/5!  because we do not need the rest of 5 digits arrangement as we wont be using them.

The above would get me the possible arrangements for "3 hundreds" like 332, 304, 398 etc... which are  42 .

Thus i now apply the same principle for rest of 4,5,7 and 9 "hundreds"

or essentially  42*5=210

However considering your 3rd point:

When I apply combination on the tens and units places in each hundred, I am not considering 3  4   4  or 3  9   9  arrangement because the digit are only used once in this method and therefore I am falling 35 arrangements short?

Sajjad1093 · Answer

my bad i didnt write clearly.

so i have assumed the following:

1) I considered "different 3-digit numbers" is all digits being different. like 334 or 332 not allowed. 302 and 320 are allowed becuase each of the 3 digit is different. But now you have stated that digits can be repeated, its the figure that needs to be different.

2) So even considering the above correction i have done  7!  because considering the limitation on the digits to be used which is 1,6,8 being off limit which leaves us with 7 digits.

I have also applied permutations and not combinations. As i have calculated 370 and 307 as unique arrangement

Like for 3 _ _ i have calculated  7!/5!  because we do not need the rest of 5 digits arrangement as we wont be using them.

The above would get me the possible arrangements for "3 hundreds" like 332, 304, 398 etc... which are  42 .

Thus i now apply the same principle for rest of 4,5,7 and 9 "hundreds"

or essentially  42*5=210

However considering your 3rd point:

When I apply combination on the tens and units places in each hundred, I am not considering 3  4   4  or 3  9   9  arrangement because the digit are only used once in this method and therefore I am falling 35 arrangements short?

EMPOWERgmatRichC · Answer

Hi Sajjad1093,

In this prompt, digits CAN be used repeatedly, so since ANY of the 7 digits can be in the "ten's spot" and ANY of the 7 digits can be in the "unit's spot", there are (7)(7) options for those two spots. However, your calculation... 7!/5! = (7)(6)... calculates that the two digits CANNOT be repeated. That's why your total is lower than it should be - there are MORE options than you are accounting for with this calculation.

GMAT assassins aren't born, they're made,
Rich

BrentGMATPrepNow · Answer

Take the task of creating the 3-digit numbers and break it into  stages .

Stage 1 : Select the first digit (hundreds digit) 
Since the first digit can be 3, 4, 5, 7 or 9, we can complete stage 1 in  5  ways

Stage 2 : Select the second digit (tens digit) 
Since the second digit can be 0, 2, 3, 4, 5, 7 or 9, we can complete stage 2 in  7  ways

Stage 3 : Select the third digit (units digit) 
Since the third digit can be 0, 2, 3, 4, 5, 7 or 9, we can complete stage 3 in  7  ways

By the Fundamental Counting Principle (FCP), we can complete all 3 stages (and thus create a 3-digit number) in  (5)(7)(7)  ways (=   2445   ways)

Answer: B

Note: the FCP can be used to solve the MAJORITY of counting questions on the GMAT. So, be sure to learn it.

RELATED VIDEOS
 https://www.youtube.com/watch?v=w5d5bvCRyd4

https://www.youtube.com/watch?v=_PFBp2BoMUY

ScottTargetTestPrep · Answer

The first (or hundreds) digit has 5 choices (digits 3 to 9, excluding 6 and 8). Each of the second (or tens) and the third (or units) digits has 7 choices (digits 0 to 9, excluding 1, 6 and 8). Therefore, the number of 3-digit numbers greater than 299 that do not contain the digits 1, 6, or 8 is

5 x 7 x 7 = 245

Answer: B

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How many different 3-digit numbers are greater than 299 and do not con

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