Question

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 $

Answer 1

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)!}} $ .

Complete step-by-step answer:
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!}} $ .
On further solving we can write
 $ \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 $ .
So, the correct 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.