How many three-digit positive integers have three different

Question

How many three-digit positive integers have three different digits such that the median of the three digits is the tens digit?

(A) 168
(B) 180
(C) 192
(D) 204
(E) 232

rajdeep212003 · Answer

Very interesting sum

Think of a 3-digit number:

_ _ _
 a b c

Here b (tens digit) is the median.

So one digit must be smaller than b and one must be greater than b.

So:
a > b > c OR c > b > a

---

Fix b.

Digits smaller than b = b 
(includes 0)

Digits greater than b = (9 − b)

So total pairs = b(9 − b)

---

Now b can be from 1 to 8:

b = 1 → 1×8 
b = 2 → 2×7 
b = 3 → 3×6 
b = 4 → 4×5 
b = 5 → 5×4 
b = 6 → 6×3 
b = 7 → 7×2 
b = 8 → 8×1

Sum = 120

---

Each pair can be arranged in 2 ways:
(a,b,c) and (c,b,a)

Total = 120 × 2 = 240

---

Now remove invalid cases.

Invalid happens when a = 0 (hundreds digit cannot be 0).

This occurs when:
- 0 is chosen as the smaller digit 
- and placed in position “a”

For each b, number of such cases = (9 − b)

So total invalid = 
8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36

---

Final = 240 − 36 = 204

Final Answer: 204

Hope this helps—feel free to upvote if useful. 
If you have similar questions, tag me—I’ll try to help.

— Rajdeep

kevincan · Answer

We could also think in terms of two cases:

Case one: no digit is equal to zero

The three digits will be three of the nine digits from 1 to 9. The hundreds digit will be either the greatest or the smallest of the three. The tens digit will be the median of the three .

9C3 * 2=168

Case two: one of the digits is equal to zero

If 0 is one of the digits, the other two digits are chosen from the remaining nine digits. The higher of the two digit chosen will be the hundreds digit.

9C2 = 36

The answer is 168+36 = 204

rajdeep212003 · Answer

that's also true thank u for sharing this it makes it even easier

HarshavardhanR · Answer

We need to find the number of three-digit positive integers, all three digits being different, where the tens digit is the median.

What does our number look like?

"abc"

Where, either ->

(1) a < b < c (ascending order)

or

(2) a > b > c (descending order)

Case 1: One fringe case that immediately comes to mind: can the number abc contain a "0"? Let's explore the possibilities.

a) 0bc --- not a 3-digit number. No need to worry about this.
b) a0c --- this case is not possible. Because we are dealing with digits (nonnegative numbers). There is no number lesser than 0 that can put at "a" or "c". 0 cannot be the middle value. No need to worry about this.

c) ab0 --- Here, there are clearly legit cases, such as 210, 540, etc (descending order). We need to count all possible numbers here.

- How many numbers, in total, exist in the format "ab0" (where a and b are different numbers)? 9 x 8 = 72
- For any 2 different numbers a and b, how many arrangements are possible? 2! = 2
- Out of these, in how many arrangements are a and b in descending order (a>b, which is what we need)? 1
- So, out of the total numbers above, in how many numbers will we see descending order of ab (a>b)? 1/2! x 72 = 36 (i.e. exactly half of the 72 numbers will have a>b and the other half will have a<b).

Case 1: 36

Case 2: a, b, and c are all non-0 digits.

- How many numbers, in total, exist in the format "abc" (where a, b, and c are different numbers)? 9 x 8 x 7
- For any 3 different numbers a,b, and c, picked, how many arrangements are possible? 3! = 6
- But out of these arrangements, how many will have ascending order a<b<c? Exactly 1

What this means is that exactly 1/6 of the total numbers (9x8x7) will have a<b<c.

Similarly

- How many arrangements, out of the 6 possible arrangements of (abc) will have proper descending order (a>b>c)? Again, Exactly 1.

What this means is that exactly 1/6 of the total numbers (9x8x7) will have a>b>c.

So, in total, out of 9x8x7 total numbers, we need to count (1/6+1/6) = 2/6 = 1/3 of the numbers.

Case 2: 1/3 x (9x8x7) = 168

Final Answer: 36 + 168 = 204. Choice D.

___
Harsha

How many three-digit positive integers have three different

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