Last visit was: 27 Apr 2026, 16:59 It is currently 27 Apr 2026, 16:59
Home
Messages
Tests
Schools
Events
Close
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track
Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
User avatar
Probus
User avatar
Joined: 10 Apr 2018
Last visit: 22 May 2020
Posts: 178
Own Kudos:
563
 [13]
Given Kudos: 115
Location: United States (NC)
Posts: 178
Kudos: 563
 [13]
How many n-digit numbers are there in which only digits used are 1 to 19 Sep 2018, 13:41
Kudos
Add Kudos
13
Bookmarks
Bookmark this Post
00:00
Start Timer
Pause Timer
Resume Timer
Show Answer
Hide
Show
History
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

45% (medium)

Question Stats:

65% (01:43) correct 35% (01:33) wrong based on 210 sessions

History

Date
Time
Result
Not Attempted Yet
How many n-digit numbers are there in which only digits used are 1 to 9 (excluding 0) and no two consecutive digits are same?

(A) \(9 * 8^n\)
(B) \(9 * 8^{n-1}\)
(C) \(9^{n-1}\)
(D) \(9^n\)
(E) \(8^n\)
ShowHide Answer
Official Answer
B
User avatar
BrentGMATPrepNow
User avatar
Major Poster
Joined: 12 Sep 2015
Last visit: 31 Oct 2025
Posts: 6,733
Own Kudos:
36,474
 [2]
Given Kudos: 799
Location: Canada
Expert
Expert reply
Posts: 6,733
Kudos: 36,474
 [2]
How many n-digit numbers are there in which only digits used are 1 to 19 Sep 2018, 15:27
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Probus
How many n-digit numbers are there in which only digits used are 1 to 9 (excluding 0) and no two consecutive digits are same?

(A)\(9 * 8^n\)
(B)\(9 * 8^{n-1}\)
(C)\(9^{n-1}\)
(D)\(9^n\)
(E)\(8^n\)
Show more

A quick approach is to test some values

For example, let's see what happens with n = 1
So, the question becomes "How many 1-digit numbers are there in which only digits used are 1 to 9 (excluding 0) and no two consecutive digits are same?"
Well, there are 9 such numbers: 1, 2, 3, 4, 5, 6, 7, 8, and 9
So, when n = 1 the answer to the question is 9

Now plug n = 1 into each answer choices and see which ones yield an output of 9
We get:
(A) 9 x 8^1 = 72. No good. We want an output of 9. ELIMINATE
(B) 9 x 8^(1-1) = 9. Great! Keep.
(C) 9^(1-1) = 1. No good. We want an output of 9. ELIMINATE
(D) 9^1 = = 9. Great! Keep.
(E) 8^1 = 8. No good. We want an output of 9. ELIMINATE

We're left with just B and D
So, we must test another value of n

Let's see what happens with n = 2
So, the question becomes "How many 2-digit numbers are there in which only digits used are 1 to 9 (excluding 0) and no two consecutive digits are same?"
Well, there are 9 options for the first digit
And there are 8 options for the second digit
So, the total number of integers that meet the give conditions = 9 x 8 = 72
So, when n = 2 the answer to the question is 72

Now plug n = 2 into each answer choices and see which ones yield an output of 72
We get:

(B) 9 x 8^(2-1) = 72. PERFECT!!
(D) 9^2 = 81. No good. We want an output of 72. ELIMINATE

Answer: B

Cheers,
Brent
User avatar
Probus
User avatar
Joined: 10 Apr 2018
Last visit: 22 May 2020
Posts: 178
Own Kudos:
563
 [1]
Given Kudos: 115
Location: United States (NC)
Posts: 178
Kudos: 563
 [1]
How many n-digit numbers are there in which only digits used are 1 to 20 Sep 2018, 14:43
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
GMATPrepNow
Probus
How many n-digit numbers are there in which only digits used are 1 to 9 (excluding 0) and no two consecutive digits are same?

(A)\(9 * 8^n\)
(B)\(9 * 8^{n-1}\)
(C)\(9^{n-1}\)
(D)\(9^n\)
(E)\(8^n\)

A quick approach is to test some values

For example, let's see what happens with n = 1
So, the question becomes "How many 1-digit numbers are there in which only digits used are 1 to 9 (excluding 0) and no two consecutive digits are same?"
Well, there are 9 such numbers: 1, 2, 3, 4, 5, 6, 7, 8, and 9
So, when n = 1 the answer to the question is 9

Now plug n = 1 into each answer choices and see which ones yield an output of 9
We get:
(A) 9 x 8^1 = 72. No good. We want an output of 9. ELIMINATE
(B) 9 x 8^(1-1) = 9. Great! Keep.
(C) 9^(1-1) = 1. No good. We want an output of 9. ELIMINATE
(D) 9^1 = = 9. Great! Keep.
(E) 8^1 = 8. No good. We want an output of 9. ELIMINATE

We're left with just B and D
So, we must test another value of n

Let's see what happens with n = 2
So, the question becomes "How many 2-digit numbers are there in which only digits used are 1 to 9 (excluding 0) and no two consecutive digits are same?"
Well, there are 9 options for the first digit
And there are 8 options for the second digit
So, the total number of integers that meet the give conditions = 9 x 8 = 72
So, when n = 2 the answer to the question is 72

Now plug n = 2 into each answer choices and see which ones yield an output of 72
We get:

(B) 9 x 8^(2-1) = 72. PERFECT!!
(D) 9^2 = 81. No good. We want an output of 72. ELIMINATE

Answer: B

Cheers,
Brent
Show more

Hello All ,

BrentGMATPrepNow

Brent, as always your methodical approach is brilliant. Just an observation most of the replies that i have seen from you are always pure method and simple concept application. I am sure there is a takeaway for all of us to not find shortcuts / formula for every type of questions rather try tweaking the basic concept and see how it can help.

Having said that the solution for this question is like this

Say we have n-digits then
- - - - - - - - - - - - - - - - upto n

First digit from left ( marked in green) cannot be 0 . So first place can be filled by any of 9 digits. So 9 ways .
Second digit from left marked in black can any 8 digits . So 8 ways
Third digit from left can again be filled by any of 8 digits. so 8 ways
.
.
.
.
n the digit marked in red can be filled by any of the 8 digits so 8 ways

So we have first space filled in 9 ways and (n-1) spaces filled 8 ways each

So we have
\(9* 8^{(n-1)}\)

Probus
User avatar
EgmatQuantExpert
User avatar
e-GMAT Representative
Joined: 04 Jan 2015
Last visit: 02 Apr 2024
Posts: 3,657
Own Kudos:
20,897
Given Kudos: 165
Expert
Expert reply
Posts: 3,657
Kudos: 20,897
How many n-digit numbers are there in which only digits used are 1 to 30 Jul 2020, 07:14
Kudos
Add Kudos
Bookmarks
Bookmark this Post

Solution


Given

    • n-digit numbers are formed using only 1 to 9.
    • Question implies that repetition is allowed.

To Find

    • How many such numbers are possible.

Approach and Working Out

    • Number of ways to select the first digit is 9.
      o For next digit we cannot use the first one and hence only 8 choices.
      o For the third digit, we cannot use the second one and hence only 8 choices.
    • For every subsequent digit, there will be 8 choices.
    • Total number of ways = 9 × 8 × 8 × 8 …. Till n digits.
      o 9 × \(8^{n-1}\) ways

Correct Answer: Option B
User avatar
sarthak1701
User avatar
Joined: 11 Sep 2024
Last visit: 14 Apr 2026
Posts: 108
Own Kudos:
65
Given Kudos: 18
GMAT Focus 1: 575 Q77 V81 DI78
GMAT Focus 1: 575 Q77 V81 DI78
Posts: 108
Kudos: 65
How many n-digit numbers are there in which only digits used are 1 to 22 Feb 2025, 06:08
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Let n = 2

We cannot have a 2-digit number with 0 or with repeating digits

Among 10s we have 8 such numbers = 12, 13, 14, 15, 16, 17, 18 and 19

This pattern would hold true for 20s, 30s up until 90s, the last relevant digit being 98

Therefore we have 8*9 = 72 such numbers. Only option B satisfies this

9 * 8^(n-1) = 9 * 8 = 72

Ans B
Moderators:
Bunuel
Math Expert
109929 posts
Krunaal
Tuck School Moderator
852 posts