Mathematics - Permutation & Combination Question with Solution | TestHub

The given numbers are $1,2,3,4$ and sum of digits should be $12$.

Case $1: \underline{4} \underline{3} \underline{1} \underline{1} \underline{1} \underline{1} \underline{1}$ ,Here the sum of the digits is $12$.

The number of ways arranging them is $=\frac{7!}{5!}=42$.

Case $2: \underline{4} \underline{2} \underline{2} \underline{1} \underline{1} \underline{1} \underline{1}$.

The number of ways arranging them is $=\frac{7!}{2!4!}=105$

Case $3: \underline{3} \underline{2} \underline{2} \underline{2} \underline{1} \underline{1} \underline{1}$.

The number of ways arranging them is $=\frac{7!}{3!3!}=140$

Case $4: \underline{3} \underline{3} \underline{2} \underline{1} \underline{1} \underline{1} \underline{1}$.

The number of ways arranging them is $=\frac{7!}{2!4!}=105$

Case $5: \underline{2} \underline{2} \underline{2} \underline{2} \underline{2} \underline{1} \underline{1}$

The number of ways arranging them is $=\frac{7!}{5!2!}=21$

The total number of ways are $=42+105+140+105+21$

$=413$

Therefore, the required answer is $413$.