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## Permutation and combination

##### This topic has expert replies
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### Permutation and combination

by parveen110 » Tue Feb 04, 2014 5:22 am
The number of natural numbers of two or more than two digits in which digits from left to right are in increasing order is
a.127
b.128
c.502
d.501
e.512

OA: C

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by GMATGuruNY » Tue Feb 04, 2014 8:13 am
parveen110 wrote:The number of natural numbers of two or more than two digits in which digits from left to right are in increasing order is
a.127
b.128
c.502
d.501
e.512

OA: C
This problem seems a bit too convoluted for the GMAT.
Also, the GMAT would say positive integers rather than natural numbers.
That said, the problem does provide some worthwhile take-aways, so here's one approach:

For the digits to increase from left to right, they must be DISTINCT.
For each combination of distinct digits, there is only ONE acceptable ordering: from least to greatest.
For example, only one viable integer can be formed if the selected digits are 7, 1 and 3:
137.
Thus, for the each of the following cases, we need to count how many COMBINATIONS of distinct digits can be formed, since each combination of distinct digits will yield exactly one viable integer.

Note the following constraint:
None of the digits in the following cases can be 0.
The reason:
Since the digits must increase from left to right, any combination that includes 0 must place 0 in the leftmost position -- not possible, since the leftmost digit of a positive integer must be at least 1.
Thus, all of the following cases must be formed from the following pool of digits:
1, 2, 3, 4, 5, 6, 7, 8, 9.

We should also note the following:
nCr = nC(n-r).
To illustrate:
6C5 = 6C(6-5) = 6C1.
From 6 options, the number of ways to choose 5 = 6C5 = (6*5*4*3*2)/(5*4*3*2*1) = 6.
From 6 options, the number of ways to choose 1 = 6C1 = 6.
Thus:
6C5 = 6C1.

2-digit integers:
Number of ways to choose 2 distinct integers from 9 options = 9C2 = (9*8)/(2*1) = 36.

3-digit integers:
Number of ways to choose 3 distinct integers from 9 options = 9C3 = (9*8*7)/(3*2*1) = 84.

4-digit integers:
Number of ways to choose 4 distinct integers from 9 options = 9C4 = (9*8*7*6)/(4*3*2*1) = 126.

5-digit integers:
Number of ways to choose 5 distinct integers from 9 options = 9C5 = 9C4 = 126.

6-digit integers:
Number of ways to choose 6 distinct integers from 9 options = 9C6 = 9C3 = 84.

7-digit integers:
Number of ways to choose 7 distinct integers from 9 options = 9C7 = 9C2 = 36.

8-digit integers:
Number of ways to choose 8 distinct integers from 9 options = 9C8 = 9C1 = 9.

9-digit integers:
Here, there is only 1 option:
123456789.

Total options = 36+84+126+126+84+36+9+1 = 502.

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by Brent@GMATPrepNow » Tue Feb 04, 2014 8:14 am
parveen110 wrote:The number of natural numbers of two or more than two digits in which digits from left to right are in increasing order is
a.127
b.128
c.502
d.501
e.512
Consider the arrangement 123456789

Now PRETEND that each digit has a lightbulb inside of it, and each lightbulb is either on or off.
If the lightbulb is ON, we can see that digit, if that lightbulb is OFF, we cannot see the digit.
So, for example, if we turn on the lightbulbs for digits 2, 5, 7 and 8, and turn off the remaining lightbulbs, we see the resulting number 2578
Notice that the digits are nicely arranged in ascending order. In fact, with this ON/OFF setup, every resulting number will have its digits arranged in ascending order.

So, all we need to do now is count the number different ways to turn on and turn off the indivudual lightbulbs.
Let's take the task of turning the lightbulbs on and off, and break it into stages.

Stage 1: Determine the lightbulb situation for digit 1
There are only two possibilities here: the lightbulb is ON or it's OFF
So, we can complete stage 1 in 2 ways

Stage 2: Determine the lightbulb situation for digit 2
There are only two possibilities here: the lightbulb is ON or it's OFF
So, we can complete stage 2 in 2 ways

Stage 3: Determine the lightbulb situation for digit 3
There are only two possibilities here: the lightbulb is ON or it's OFF
So, we can complete stage 3 in 2 ways
.
.
.
Stage 9: Determine the lightbulb situation for digit 9
There are only two possibilities here: the lightbulb is ON or it's OFF
So, we can complete stage 9 in 2 ways

By the Fundamental Counting Principle (FCP), we can complete all 9 stages (and thus create a number with its digits in ascending order) in (2)(2)(2)(2)(2)(2)(2)(2)(2) ways (= 512 ways

IMPORTANT: Among the 512 possibilities, we have accidentally included some scenarios that don't meet the conditions in the question.
For example, among the 512 possibilities, we have accidentally included 1 scenario where ALL of the lightbulbs are off. In this case, there is NO resulting number.
Likewise, the question tells us that we must have 2 or more digits, and among the 512 possibilities, we have accidentally included 9 scenarios where EXACTLY ONE lightbulb is on (resulting in 1, 2, 3, 4,...8, and 9).

So, the number of numbers that meet the given conditions = 512 - 1 - 9
= 502
= C

Cheers,
Brent

Last edited by Brent@GMATPrepNow on Thu Feb 13, 2014 6:30 am, edited 1 time in total.
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by parveen110 » Thu Feb 13, 2014 4:01 am
Brent@GMATPrepNow wrote:
parveen110 wrote:The number of natural numbers of two or more than two digits in which digits from left to right are in increasing order is
a.127
b.128
c.502
d.501
e.512
Consider the arrangement 123456789

Now PRETEND that each digit has a lightbulb inside of it, and each lightbulb is either on or off.
If the lightbulb is ON, we can see that digit, if that lightbulb is OFF, we cannot see the digit.
So, for example, if we turn on the lightbulbs for digits 2, 5, 7 and 8, and turn off the remaining lightbulbs, we see the resulting number 2578
Notice that the digits are nicely arranged in ascending order. In fact, with this ON/OFF setup, every resulting number will have its digits arranged in ascending order.

So, all we need to do now is count the number different ways to turn on and turn off the indivudual lightbulbs.
Let's take the task of turning the lightbulbs on and off, and break it into stages.

Stage 1: Determine the lightbulb situation for digit 1
There are only two possibilities here: the lightbulb is ON or it's OFF
So, we can complete stage 1 in 2 ways

Stage 2: Determine the lightbulb situation for digit 2
There are only two possibilities here: the lightbulb is ON or it's OFF
So, we can complete stage 2 in 2 ways

Stage 3: Determine the lightbulb situation for digit 3
There are only two possibilities here: the lightbulb is ON or it's OFF
So, we can complete stage 3 in 2 ways
.
.
.
Stage 9: Determine the lightbulb situation for digit 9
There are only two possibilities here: the lightbulb is ON or it's OFF
So, we can complete stage 9 in 2 ways

By the Fundamental Counting Principle (FCP), we can complete all 9 stages (and thus create a number with its digits in ascending order) in (2)(2)(2)(2)(2)(2)(2)(2)(2) ways (= 512 ways

IMPORTANT: Among the 512 possibilities, we have accidentally included some scenarios that don't meet the conditions in the question.
For example, among the 512 possibilities, we have accidentally included 1 scenario where ALL of the lightbulbs are off. In this case, there is NO resulting number.
Likewise, the question tells us that we must have 2 or more digits, and among the 512 possibilities, we have accidentally included 9 scenarios where EXACTLY ONE lightbulb is off (resulting in 1, 2, 3, 4,...8, and 9).

So, the number of numbers that meet the given conditions = 512 - 1 - 9
= 502
= C

Cheers,
Brent

Hi Brent,

Brilliant approach,as usual..but in the solution you suggested, are you meaning to say that "we have accidentally included 9 scenarios where EXACTLY ONE lightbulb is 'ON' (resulting in 1, 2, 3, 4,...8, and 9)?? Thanks.

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by Brent@GMATPrepNow » Thu Feb 13, 2014 6:32 am
parveen110 wrote:
Hi Brent,

Brilliant approach,as usual..but in the solution you suggested, are you meaning to say that "we have accidentally included 9 scenarios where EXACTLY ONE lightbulb is 'ON' (resulting in 1, 2, 3, 4,...8, and 9)?? Thanks.
Good catch, parveen110, thanks!
I've edited my solution accordingly.

Cheers,
Brent
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by nikhilgmat31 » Thu Oct 08, 2015 1:24 am
This question seems too tough for GMAT.

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by vinsmba » Sun May 21, 2017 12:36 pm
just wondering how to formulate that digit..i am seeing an explanation but can't understand the 'ask' here. n >= 2 digits does it mean 2 digit or 3 or 4,5,6,7,..keeps going on. what is the limitation that we apply or how best can this be broken? why should we consider 123456789 and not any repetitions witin it. (looks like old post but just stuck with this understanding)

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by Brent@GMATPrepNow » Mon May 22, 2017 6:00 am
vinsmba wrote:just wondering how to formulate that digit..i am seeing an explanation but can't understand the 'ask' here. n >= 2 digits does it mean 2 digit or 3 or 4,5,6,7,..keeps going on. what is the limitation that we apply or how best can this be broken? why should we consider 123456789 and not any repetitions witin it. (looks like old post but just stuck with this understanding)

Aside: "of two or more than two digits" means "two or more digits" (if that's what you're asking)

Cheers,
Brent
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by Brent@GMATPrepNow » Mon May 22, 2017 6:01 am
nikhilgmat31 wrote:This question seems too tough for GMAT.
Quite possibly.
If it is within the scope of the GMAT, it's a 750+ level question.

Cheers,
Brent
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