The number of five digit numbers that can be formed by using digits 1, 2, 3 only, such that three digits of the formed number are identical, is

When exactly three digits are identical and the remaining two are different, then the number of such numbers

$\left({ }^3{C}_1\right)\left(\frac{5!}{3!}\right)=60$

When three digits are identical and the remaining two are also identical, then the number of ways

$=\left({ }^3{C}_2\right)\left(2\right)\left(\frac{5!}{3!2!}\right)=60$

Thus, the number of such numbers$= 120.$