A four-digit number is formed by using only the digits $1$, $2$ and $3$ such that both $2$ and $3$ appear at least once. The number of all such four-digit numbers is

We need to form four-digit numbers using only the digits 1, 2, and 3, with one important condition: both 2 and 3 must appear at least once in each number.

Think of it this way: We have 4 positions to fill, and we're not allowed to use numbers that contain only 1s and 2s, or only 1s and 3s, or only 1s.

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Since both 2 and 3 must appear at least once, let's systematically list all possible ways to distribute these digits across 4 positions.

When we have repeated elements, we use the formula for permutations with repetition:

$\dfrac{n!}{n_1! \times n_2! \times n_3! \times ...}$

where $n$ is total positions and $n_1, n_2, n_3...$ are frequencies of each repeated element.

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Case 1: (2, 3, 1, 1) - One 2, one 3, two 1s

We need to arrange the digits 2, 3, 1, 1 in 4 positions.

$\dfrac{4!}{2!} = \dfrac{24}{2}$

$= 12$ arrangements

We divide by 2! because the digit 1 appears twice.

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Case 2: (2, 3, 1, 2) - Two 2s, one 3, one 1

We need to arrange 2, 2, 3, 1 in 4 positions.

$\dfrac{4!}{2!} = 12$ arrangements

We divide by 2! because the digit 2 appears twice.

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Case 3: (2, 3, 1, 3) - One 2, two 3s, one 1

We need to arrange 2, 3, 3, 1 in 4 positions.

$\dfrac{4!}{2!} = 12$ arrangements

We divide by 2! because the digit 3 appears twice.

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Case 4: (2, 3, 2, 2) - Three 2s, one 3

We need to arrange 2, 2, 2, 3 in 4 positions.

$\dfrac{4!}{3!} = \dfrac{24}{6}$

$= 4$ arrangements

We divide by 3! because the digit 2 appears three times.

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Case 5: (2, 3, 2, 3) - Two 2s, two 3s

We need to arrange 2, 2, 3, 3 in 4 positions.

$\dfrac{4!}{2! \times 2!} = \dfrac{24}{4}$

$= 6$ arrangements

We divide by 2! twice because both 2 and 3 appear twice each.

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Case 6: (2, 3, 3, 3) - One 2, three 3s

We need to arrange 2, 3, 3, 3 in 4 positions.

$\dfrac{4!}{3!} = 4$ arrangements

We divide by 3! because the digit 3 appears three times.

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Adding up all cases:

$12 + 12 + 12 + 4 + 6 + 4 = 50$

This systematic approach ensures we don't miss any valid combinations while avoiding double-counting. Each case represents a unique way to distribute the digits 1, 2, and 3 across four positions while satisfying our constraint.

The answer is 50.