Answer

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Hint: We are supposed to form a 6-digit natural number using the 4 digits 1,2,3,4, think of all the possible cases before starting the solution.

We have a total of $4$ digits using which we have to create a $6$ digit number, the $4$ digits are fixed but the only change that can happen is in the remaining two digits, there can be two cases, one where both digits are same and the other case is that both the digits are distinct.

Therefore,

Case 1: Digits are the same.

Out of the six digits, four are fixed and the other two digits can be any of the four digits,

Therefore,

$ \Rightarrow {}^4{C_1} \times \dfrac{{6!}}{{3!}}$

Case 2: Digits are distinct

$ \Rightarrow {}^4{C_2} \times \dfrac{{6!}}{{2! \times 2!}}$

Total $ \Rightarrow 6!\left( {23 + 32} \right)$

$ \Rightarrow 6! \times 136$

$ \Rightarrow 1560$

Answer $ \Rightarrow 1560$ numbers can be formed.

Note: In this question, there were only 2 cases, but there are questions which might have more than 2 cases make sure of thinking about all the cases, before solving the question.

We have a total of $4$ digits using which we have to create a $6$ digit number, the $4$ digits are fixed but the only change that can happen is in the remaining two digits, there can be two cases, one where both digits are same and the other case is that both the digits are distinct.

Therefore,

Case 1: Digits are the same.

Out of the six digits, four are fixed and the other two digits can be any of the four digits,

Therefore,

$ \Rightarrow {}^4{C_1} \times \dfrac{{6!}}{{3!}}$

Case 2: Digits are distinct

$ \Rightarrow {}^4{C_2} \times \dfrac{{6!}}{{2! \times 2!}}$

Total $ \Rightarrow 6!\left( {23 + 32} \right)$

$ \Rightarrow 6! \times 136$

$ \Rightarrow 1560$

Answer $ \Rightarrow 1560$ numbers can be formed.

Note: In this question, there were only 2 cases, but there are questions which might have more than 2 cases make sure of thinking about all the cases, before solving the question.

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