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# How many n-digit numbers are there in which only digits used are 1 to

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Manager
Joined: 10 Apr 2018
Posts: 180
How many n-digit numbers are there in which only digits used are 1 to  [#permalink]

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19 Sep 2018, 12:41
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00:00

Difficulty:

35% (medium)

Question Stats:

66% (01:29) correct 34% (01:17) wrong based on 32 sessions

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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) $$8^n$$
CEO
Joined: 11 Sep 2015
Posts: 3341
Re: How many n-digit numbers are there in which only digits used are 1 to  [#permalink]

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19 Sep 2018, 14:27
2
Top Contributor
Probus wrote:
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

Cheers,
Brent
_________________

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Manager
Joined: 10 Apr 2018
Posts: 180
How many n-digit numbers are there in which only digits used are 1 to  [#permalink]

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20 Sep 2018, 13:43
GMATPrepNow wrote:
Probus wrote:
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

Cheers,
Brent

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
How many n-digit numbers are there in which only digits used are 1 to &nbs [#permalink] 20 Sep 2018, 13:43
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