Class 11

Math

Algebra

Permutations and Combinations

Three digit numbers in which the middle one is a perfect square are formed using the digits $1$ to $9$. Their sum is

- $134055$
- $270540$
- $170055$
- $270504$

Let $ABC$ be the number

$Aϵ{1,2,3.....9}$ $Bϵ{1,4,9}$

$Cϵ{0,1,2,.....9}$

Each possible choice for $B$ occurs. $81$ times $[9×9]$

Each possible choice for $A$ & $C$ occurs $27$ times $[3×9]$

$∴$ Sum of all numbers

$=27×100×(1+2+3+.....+9)+81×10(1+4+9)+27×1×(1+2+3.....9)$

$=27×45×100+81×10×14+27×1×45$

$=134055$

Hence, this is the answer.