How many three-digit positive integers are there such that the tens

Question

How many three-digit positive integers are there such that the tens digit is smaller than the hundreds digit but greater than the units digit?

(A) 60
(B) 72
(C) 96
(D) 120
(E) 144

poojaarora1818 · Answer

Solution:

We have to select three distinct digits so that the Hundreds digit> Tens digit> Units digit.

This can be solved by applying the combination method:

Case 1. We have select any three distinct digits among these nine digits for hundreds digit except zero. (0,1,2,3,4,5,6,7,8,9).

9C3 = \frac{ 9*8*7}{3!}  = 84 ways.

Case 2. We have to select any two distinct digits for tens digit including zero but excluding the digit selected for the hundreds place. (0,1,2,3,4,5,6,7,8,9).

9C2 =  \frac{9*8}{2!}  = 36 ways.

By adding Case 1 and 2 we get 120 ways.

Hence, Option D

Bunuel · Answer

The three digits must be distinct (hundreds digit > tens digit > units digit), so we are just choosing 3 distinct digits from the 10 digits (0, 1, 2, ..., 9). For each such triplet, exactly one arrangement satisfies the required order (hundreds digit > tens digit > units digit). Therefore, the number of such integers is simply 10C3, which is 120.

Answer: D.

Answer: D.

sujoykrdatta · Answer

We have: H>T>U
Since the least digit is the units digit, we can use the digit 0 (if used, it will be assigned to U)

So essentially we need to select 3 digits from 10 digits in 10C3 = 120 ways (for every selection there is only one way of forming the number)

Answer D

crimson_noise · Answer

Where the logic breaks

You split into:

Case 1: 9C3 = 84  
  Case 2: 9C2 = 36  
  Then added: 84 + 36 = 120

But the issue is: 
  These two cases are not logically defined or mutually exclusive.

Case 1 says: choose 3 digits from 1–9.  
  Case 2 says: choose 2 digits including zero but excluding hundreds digit.

This double-counts conceptually and does not correctly represent how digits are placed.

Also: 
 
  You are not properly assigning digits to places.  
  You are mixing selection logic for hundreds and tens digits incorrectly.  
  The addition of 84 and 36 is not justified combinatorially.

The fact that it equals 120 is pure coincidence.

poojaarora1818 · Answer

Thank you so much for pinpointing my mistakes. As I am not a maths person, this thing is quite obvious in my case. I could realize my mistakes after viewing the expert's solutions posted above.

Thank you!

Adit_ · Answer

Woah I just realised we dont even need to consider special cases of 0 in hundreds position because as soon as we choose the 3 digits they could automatically be arranged in a way where 0 is not in hundreds position since only 1 such setup is possible for any 3 digits, simply choosing any random 3 is more than enough.
Very insightful method!

How many three-digit positive integers are there such that the tens

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How many three-digit positive integers are there such that the tens

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