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How many positive numbers with two or more digits can be formed with
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03 Apr 2020, 05:38
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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
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Re: How many positive numbers with two or more digits can be formed with
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03 Apr 2020, 06:31
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 Questionspossible no 9c2+9c3+9c4+9c5+9c6+9c7+9c8+9c9 => 36+84+126+126+84+36+9+1 =>502 IMOC



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Re: How many positive numbers with two or more digits can be formed with
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04 Apr 2020, 00:41
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 Questionspossible no 9c2+9c3+9c4+9c5+9c6+9c7+9c8+9c9 => 36+84+126+126+84+36+9+1 =>502 IMOC Hi, Could you please elaborate?



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Re: How many positive numbers with two or more digits can be formed with
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05 Apr 2020, 17:22
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)



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Re: How many positive numbers with two or more digits can be formed with
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05 Apr 2020, 20:42
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
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Re: How many positive numbers with two or more digits can be formed with
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05 Apr 2020, 20:55
2^9(9c0+9c1) 512 10=502 Answer is C Remember 9c0 +9c1 +9c9 =2^9
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Re: How many positive numbers with two or more digits can be formed with
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13 Apr 2020, 00:24
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
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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
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