In how many ways can we write a 3-digit number ABC, where A

Question

In how many ways can we write a 3-digit number ABC, where A<B<C and A, B and C are digits?
(A) 7
(B) 9
(C) 75
(D) 84
(E) 120

Chetan's question set

CareerGeek · Answer

Note that, for every set of 3 different integers, there is only 1 possible way we can arrange them in an ascending order (or only one 3-digit number can be formed)
Eg: From 3 digits 1,2&3 only one 3-digit number 123 can be formed which is ascending.

Also, none of the digits can be 0, since for ascending order, 0 has to be always in hundreds place and it will never be a 3-digit number 
Eg: 056, 012, etc.

—> Total number of 3 digits possible = Total number of selections of 3 digits from available 9 non-zero digits (1-9) =  9c_3 = 84

Option D

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Yes2GMAT · Answer

Using this formula, how do u prevent A from being a number like 9? In this case, C and B cannot be greater than A.

Yes2GMAT · Answer

chetan2u  could u clarify this?

chetan2u · Answer

Yes2GMAT

Each selection(not arrangement) can further be used to arrange the items. But only one such arrangement will be in in drawing order, one in decreasing order and so on. We are looking for increasing order and only one will give you the selected items in increasing order.

Example: We select 9, 1 and 5. Only way in increasing order is 159 while other arrangements 195, 519, 591, 915 or 951 do not give the digits in increasing order.

Thus, number of selection is equal to number of ways in increasing order.

Now, abc and a<b<c means we are looking for increasing order. 
Digits are 0 to 9, and any selection involving 0 will give a( the smallest digit in a, b and c) as 0, making the number 0bc or a 2-digit number.

Thus, we cannot choose 0 and we are left to choose from remaining 9 digits.

Ways to select 3 out of 9 = 9C3 or  \frac{9*8*7}{3*2} =12*7=84

E

rajtare · Answer

I used a slightly long approach

let three digit position be  a  b  c

lets Fix the value of c as c=9, 
When, b=8, then a can have 7 values (since a cannot take the value of zero)
When b=7, then a can have 6 values
when b=6, then a can have 5 values
:
As we can see the pattern has started to develop, 
Thus when c's value is fixed as 9 we have 7+6+5+4+3+2+1 = 28 numbers
when we fix C's value as 8 we will have 6+5+4+3+2+1= 21 numbers
Also, when c's value is fixed as 7, we will have 5+4+3+2+1 = 15 numbers

Thus required numbers are : 28+21+15+10+6+3+1=84

k10890 · Answer

So when we take 9C3, we are essentially picking from the pot any three digits. Since we know there will be only 1 way of arranging in an ascending order, 9 will always be the third digit. The concern you're raising would be valid if we had used a permutation method instead.

Hope it helps.

In how many ways can we write a 3-digit number ABC, where A<B<C and A,

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