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Math Expert V
Joined: 02 Sep 2009
Posts: 64274
How many positive numbers with two or more digits can be formed with  [#permalink]

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Difficulty:   65% (hard)

Question Stats: 50% (02:43) correct 50% (02:19) wrong based on 30 sessions

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How many positive numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 so that in every such number, each digit is used at most once and the digits appear in the ascending order?

A. 500
B. 501
C. 502
D. 503
E. 504

Are You Up For the Challenge: 700 Level Questions

_________________
GMAT Club Legend  V
Joined: 18 Aug 2017
Posts: 6314
Location: India
Concentration: Sustainability, Marketing
GPA: 4
WE: Marketing (Energy and Utilities)
Re: How many positive numbers with two or more digits can be formed with  [#permalink]

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Bunuel wrote:
How many positive numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 so that in every such number, each digit is used at most once and the digits appear in the ascending order?

A. 500
B. 501
C. 502
D. 503
E. 504

Are You Up For the Challenge: 700 Level Questions

possible no
9c2+9c3+9c4+9c5+9c6+9c7+9c8+9c9
=> 36+84+126+126+84+36+9+1
=>502
IMOC
Intern  B
Joined: 30 Sep 2019
Posts: 30
Re: How many positive numbers with two or more digits can be formed with  [#permalink]

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Archit3110 wrote:
Bunuel wrote:
How many positive numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 so that in every such number, each digit is used at most once and the digits appear in the ascending order?

A. 500
B. 501
C. 502
D. 503
E. 504

Are You Up For the Challenge: 700 Level Questions

possible no
9c2+9c3+9c4+9c5+9c6+9c7+9c8+9c9
=> 36+84+126+126+84+36+9+1
=>502
IMOC

Hi,
Manager  S
Joined: 30 Jun 2019
Posts: 209
Re: How many positive numbers with two or more digits can be formed with  [#permalink]

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1
9c2+9c3+9c4+9c5+9c6+9c7+9c8+9c9

Start with two digit number, and go all the way up to a 9 digit number.
Since the numbers selected can only from 1 number (because the numbers are in ascending order), there is no need to worry about accounting for additional arragnements.

i.e
9C4 -->4567 can only form that 1 number (5467 would not work)
VP  V
Joined: 28 Jul 2016
Posts: 1020
Location: India
Concentration: Finance, Human Resources
Schools: ISB '18 (D)
GPA: 3.97
WE: Project Management (Investment Banking)
Re: How many positive numbers with two or more digits can be formed with  [#permalink]

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selecting two or more number means any number of numbers can be selected from 9 numbers
thus
total combinations are
9C2 +9 C3 +9C4+9C5+9C6+9C7+9C8+9C9

from these number arrangement will always be 1 . (sine there is only one order. suppose number of digits select are three (4,7 and 2)
thus order can only be 1: 247

thus total arrangements remain
9C2 +9 C3 +9C4+9C5+9C6+9C7+9C8+9C9
or
26+84+126+126+84+36+9+1 = 502
Thus C
_________________

Keep it simple. Keep it blank
Manager  B
Joined: 18 Dec 2017
Posts: 227
Re: How many positive numbers with two or more digits can be formed with  [#permalink]

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2^9-(9c0+9c1)
512- 10=502
Remember
9c0 +9c1 +-------9c9 =2^9

Posted from my mobile device
Manager  S
Joined: 26 Jul 2014
Posts: 110
Re: How many positive numbers with two or more digits can be formed with  [#permalink]

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How many positive numbers with two or more digits can be formed with the digits 1, 2, 3, 4, 5, 6, 7, 8, and 9 so that in every such number, each digit is used at most once and the digits appear in the ascending order?

A. 500
B. 501
C. 502
D. 503
E. 504

____________________

Initially, I was a bit confused with the concept of ascending orders. But it is pretty simple once you know.
Say, you take 5 out of 9 numbers from 1 to 9 and you want these picked number to be in the ascending order.

First you pick up 5 numbers out of 9. -- 9C5. Say you pick 1,3,2,5,6. since each number is unique, you initially have to rearrange these 5 numbers.
so the equation for doing this is 9C5 x 5!. Now, for an ascending order, you have 5 choices for the first digit place, 4 choices for the second digit and so on...
so you divide the equation above by 5!. Therefore, choosing 5 numbers out of 9 numbers with the chosen number to be in the ascending order, the equation is 9C5.

Now, back to the question.
The question is asking How many positive numbers with two or more digits can be formed?
so 2 digit number, 3 digit numbers, 4 digit numbers, and so on.

You choose 2 numbers out of 9 numbers - 9C2
You choose 3 numbers out of 9 numbers - 9C3
....
You choose 9 numbers out of 9 numbers - 9C9

Now, you add all the results together, you have 502.
Hence C. Re: How many positive numbers with two or more digits can be formed with   [#permalink] 13 Apr 2020, 00:24

# How many positive numbers with two or more digits can be formed with  