Cool Combinatorics

There $5$ numbers with all their digits equal. There $\binom{5}{4}\times 2\times 3=30$ with four equal digits.

$3^5-33=\boxed{210}$

We have essentially 3 different cases (This is the most basic approach and one that might have been used by most people) .

• Three identical digits , one identical pair

• Three identical digits , two different digits .

• Two identical pairs , one lone digit .

(i) The number of selections of 3 identical digits and one identical pair = $\binom{3}{1} \times \binom{2}{1}$ .

Therefore the total umber of numbers that can be formed is = $\binom{3}{1} \times \binom{2}{1} \times \frac{5!}{3! \times 2!}$ which equals 60 .

(ii) The number of selections of three identical digits and two different digits = $\binom{3}{1} \times \binom{2}{2}$

Therefore the number of numbers that can be formed is = $\binom{3}{1} \times \binom{2}{2} \times \frac{5!}{3!}$ which equals 60 .

(iii) The final case .The number of solutions of two identical pairs and two different digits is = $\binom{3}{1} \times \binom{1}{1}$ .

Therefore the total number of numbers that can be formed is $\binom{3}{2} \times \binom{1}{1} \times \frac{5!}{2! \times 2!}$ which equals 90 .

Hence the sum of all such numbers comes out to be $60 + 60 + 90 = 210$ .

There are 3 possible digits for each of the 5 positions so there are $3^{5}$ 5 digit integers using only the digits 1, 2, and 3. However, it is stated in the question that each digit can be used at most three times so we have to subtract the cases when there are 4 or 5 of the same digit.

Exactly 4 1s:

There are 5C4 ways of placing 4 1s and there are 2 choices for the remaining digit (namely 2 and 3) so there are 5C4 x 2 = 10 numbers containing four 1s.

It then follows that there are 10 numbers containing four 2s and 10 numbers containing four 3s. So there are 30 numbers containing the same digit 4 times.

Five repeated digits:

There are three numbers with 5 repeated digits:

11111

22222

33333

So the answer is $3^{5} -30-3 = \boxed{210}$