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

Question

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

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

Are You Up For the Challenge: 700 Level Questions

Archit3110 · Answer

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

fireagablast · Answer

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)

globaldesi · Answer

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

gurmukh · Answer

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

Posted from my mobile device

h31337u · Answer

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

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.

ettenetur · Answer

This is a bit confusing because of the verbiage. It says each digit must appear at most once which is different from “appear only once”.

If I understand the question correctly, it says that the digits from 1-9 can also not appear in the numbers.

Krunaal · Answer

It means whatever number you form, except a single digit number, the digits from 1 to 9 can only be used once in that number. Yes, you can also not use it at all, for example 12 => this does not have any digit from 3 to 9, but you need to find the  positive numbers that can be formed , so you can't say none of the 9 numbers is used.

Alternate Solution

{1, 2, 3, 4, 5, 6, 7, 8, 9}

Total number of sets that can be formed =>  2^9 = 512

Empty set => {} = 1

Single digit sets => {1} {2}.....{9} = 9

Numbers with two or more digits = 512 - 1 - 9 =  502

Answer C.

Bunuel · Answer

In this context, “at most once” means a digit can appear once or not at all, but never more than once. So yes, you’re correct: we can’t repeat digits, but we also obviously don’t have to and cannot use every digit in each number.

Now, since the digits in the number must be in ascending order, each valid number corresponds directly to choosing a subset of the digits {1,2,3,4,5,6,7,8,9}, and then writing them in ascending order.

Total subsets of 9 digits = 2^9 = 512, since each digit has two options: either it appears in the number or it does not.

Subsets of size 0 (empty set) and size 1 (single-digit numbers) are excluded, because the question asks for numbers with two or more digits.

That leaves 512 - 1 (empty set) - 9 (single-digit numbers) = 502.

So the answer is 502 (C).

amemalox · Answer

The numbers can be made in

1 • 8
1 • 1 • 7 
1 • 1 • 1 • 6 ...

ways, therefore 8! With contains 7 and the only number on the answers divisible by 7 is 504. So E but that’s wrong.

edit: now I understand that I did an error in counting the possible ways

Krunaal · Answer

I don't think so that's a valid approach to this problem, please review the solutions above.

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

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

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