How many three-digit integers greater than 150 have a tens d

Question

How many three-digit integers greater than 150 have a tens digit that is greater than the hundreds digit but less than the units digit?

A. 63
B. 66
C. 68
D. 70
E. 84

kevincan · Answer

We can consider two cases:

Case 1: The hundreds digit is 1
Case 2: the hundreds digit is greater than 1

kevincan · Answer

If h is 1 , t and u must be two distinct digits from 5 to 9 such that u is greater than t. 5C2 = 10

If h is greater than 1, h , t and u must be three distinct digits from 3 to 9, and h<t<u
8C3 = 56

10+56 = 66

ZeroSquare · Answer

There are  66  three-digit integers greater than 150 that satisfy this condition. 
 To find this number, we look for three-digit integers $abc$ where the hundreds digit ($a$), tens digit ($b$), and units digit ($c$) follow the strict inequality  $a < b < c$ , while ensuring the number is  greater than 150 . 1 ]

Breakdown by Hundreds Digit ($a$) 
 
   When $a = 1$ :

The condition $1 < b < c$ must hold.  
  To be greater than 150, the tens digit $b$ must be 5 or greater.  
  If $b = 5$, $c$ can be 6, 7, 8, 9 (4 numbers).  
  If $b = 6$, $c$ can be 7, 8, 9 (3 numbers).  
  If $b = 7$, $c$ can be 8, 9 (2 numbers).  
  If $b = 8$, $c$ can be 9 (1 number).  
   Total for $a=1$ : $4 + 3 + 2 + 1 =$  10 integers .  1 ,  2 ,  3 ]   
 
   When $a \ge 2$ :

Any three-digit number starting with 2 or more is automatically greater than 150.  
  We can find the count for each starting digit by choosing any 2 higher digits from the remaining options:  
   If $a = 2$ : Choose 2 digits from $\{3,4,5,6,7,8,9\}$ $\rightarrow \binom{7}{2} =$  21 integers   
   If $a = 3$ : Choose 2 digits from $\{4,5,6,7,8,9\}$ $\rightarrow \binom{6}{2} =$  15 integers   
   If $a = 4$ : Choose 2 digits from $\{5,6,7,8,9\}$ $\rightarrow \binom{5}{2} =$  10 integers   
   If $a = 5$ : Choose 2 digits from $\{6,7,8,9\}$ $\rightarrow \binom{4}{2} =$  6 integers   
   If $a = 6$ : Choose 2 digits from $\{7,8,9\}$ $\rightarrow \binom{3}{2} =$  3 integers   
   If $a = 7$ : Choose 2 digits from $\{8,9\}$ $\rightarrow \binom{2}{2} =$  1 integer

Final Sum 
 $\text{Total}=10+21+15+10+6+3+1=\mathbf{66}$ 
 If you want to explore more digit puzzles, let me know if you would like to  change the inequality constraints or  restrict the allowed digits

ZetaHope · Answer

But if we calculate it, the numbers with hundreds digit are 10 not 28.

How many three-digit integers greater than 150 have a tens d

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How many three-digit integers greater than 150 have a tens d

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