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# In how many ways can $3$ sovereign be given away when there are $4$ applicants and any applicant may have either $0,1,2$ or $3$ sovereigns? A. $15$ B. $20$ C. $24$ D. $48$ Verified
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Hint: As we can see that in the above question we have to use the combination because we have to select the number of way such that $3$ sovereign can be given away when there are $4$ applicants and any applicant can be given either $0,1,2$ or $3$ sovereigns. 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 $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ .

Here we have a total of $3$ sovereigns. Let us assume the non-negative integral of solutions i.e. $x,y,z,w$ .
So we can write $x + y + z + w = 3$ .
We know the formula which is $^{n + r - 1}{C_{r - 1}}$ . In this question we have $n = 3$ and $r = 4$ .
So by putting the value we have $^{3 + 4 - 1}{C_{4 - 1}}$ , it gives us $^6{C_3}$ .
Now by applying the formula of combination we can write $\dfrac{{6!}}{{3!(6 - 3)!}} \Rightarrow \dfrac{{6!}}{{3!3!}}$ .
$\dfrac{{6 \times 5 \times 4 \times 3!}}{{3 \times 2 \times 1 \times 3!}} = \dfrac{{6 \times 5 \times 4}}{{3 \times 2}}$ . It gives us value $\dfrac{{120}}{6} = 20$ .
Hence the required answer is $20$ .
Note: We should be careful about what we have to use in the solution i.e. permutation or combination. While applying the formula of combination i.e. $^n{C_r} = \dfrac{{n!}}{{r!(n - r)!}}$ in $^6{C_3}$ , by comparing we have here $n = 6,r = 3$ . 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.