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# How many $6$ digit numbers can be formed using the digits $1,2,3$ only, whose sum of digits is $10?$

Last updated date: 10th Aug 2024
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Hint:As we can see in the above question we have to use the combination because we have to choose the numbers to make six digits in such a way that the sum has to be $10$. We know that the selection of some or all objects from a given set of different objects is called combination. The formula of combination is written as $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$.

We have been given numbers: $1,2,3$. We have to form six digits from these so that the sum is $10$. Let us make the first selection i.e. $2,2,2,2,1,1$. Now we make the second selection: $(3,3,1,1,1,1)$. We can see that there can be no more selection of numbers from these given numbers.

In the first order we have $2,2,2,2,1,1$. So we have $n = 6$, We have $4$ repeated four times, so $r = 4$. Now we can put this in the formula $\dfrac{{6!}}{{4!(6 - 4)!}}$.
On solving we have,
$\dfrac{{6!}}{{4!2!}} = \dfrac{{6 \times 5 \times 4!}}{{4! \times 2 \times 1}}$.
Further simplifying we have the value: $\dfrac{{6 \times 5}}{2} = 15$.
Similarly we can solve the second selection: $(3,3,1,1,1,1)$.
So we have $n = 6$, We have $1$ repeated four times, so $r = 4$. Now we can put this in the formula $\dfrac{{6!}}{{4!(6 - 4)!}}$.
On solving we have,
$\dfrac{{6!}}{{4!2!}} = \dfrac{{6 \times 5 \times 4!}}{{4! \times 2 \times 1}}$
Further simplifying we have the value: $\dfrac{{6 \times 5}}{2} = 15$.
So we have a total two ways, we add them to get the total number i.e. $15 + 15 = 30$.

Hence the required answer is $30$.

Note:We should be careful about what we have to use in the solution i.e. permutation or combination. When we arrange objects in a definite order then we call it Permutation. The formula of Permutation is written as $^n{P_r} = \dfrac{{n!}}{{(n - r)!}}$. In permutation we know that $0 < r \leqslant n$, it should be cross checked to avoid calculation errors.