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Question: In how many ways can \(3\) sovereigns be given away when there are \(4\) applicants and any applican...

In how many ways can 33 sovereigns be given away when there are 44 applicants and any applicant may have either 0,1,20,1,2 or 33sovereigns.
A)15A)15
B)20B)20
C)24C)24
D)48D)48

Explanation

Solution

To solve this question we need to have the concept of combinations. We will be analysing the problem and will be considering 4 non integral numbers which will result into three sovereigns and then writing of the number of ways in which it should be distributed among the four applicants.

Complete step-by-step solution:
The question asks us to find the number of ways in which three sovereigns should be given to four applicants where the applicants can be given either of 0,1,20,1,2or 33 sovereigns.
The first step to solve this question is to consider four non-negative integral variables which show the number of sovereigns given to the four applicants, so consider the variables to be x,y,z,wx,y,z,w . So on writing it mathematically we get:
⇒x+y+z+w=3\Rightarrow x+y+z+w=3
The second step is to find the number of required ways in which number of sovereigns should be distributed. For calculation we will use the function of Combination. So in mathematical form we will write:
⇒(3+4−1)C(4−1)\Rightarrow {}^{\left( 3+4-1 \right)}{{C}_{\left( 4-1 \right)}}
The above formula shows the distribution of n things among r distinct persons as here we had to give the 3 sovereigns to 4 persons.
On calculating the term we get:
⇒6C3\Rightarrow {}^{6}{{C}_{3}}
To solve the above Combination we will use the formula of the combination nCr{}^{n}{{C}_{r}} as n!r!(n−r)!\dfrac{n!}{r!\left( n-r \right)!} . On applying the same formula we get:
⇒6C3=6!3!(6−3)!\Rightarrow {}^{6}{{C}_{3}}=\dfrac{6!}{3!\left( 6-3 \right)!}
⇒6C3=6!3!3!\Rightarrow {}^{6}{{C}_{3}}=\dfrac{6!}{3!3!}
On further calculation the above expression changes to:
⇒6C3=6×5×4×3!3!3!\Rightarrow {}^{6}{{C}_{3}}=\dfrac{6\times 5\times 4\times 3!}{3!3!}
Now the term 3!3! will cancel since the term is common.
⇒6C3=6×5×43×2×1\Rightarrow {}^{6}{{C}_{3}}=\dfrac{6\times 5\times 4}{3\times 2\times 1}
⇒6C3=5×4\Rightarrow {}^{6}{{C}_{3}}=5\times 4
⇒6C3=20\Rightarrow {}^{6}{{C}_{3}}=20
∴\therefore The number of ways can 33 sovereigns be given away when there are 44 applicants and any applicant may have either 0,1,20,1,2 or 33sovereigns are option B)20B)20.

Note: We need to remember the formula for the Combination nCr{}^{n}{{C}_{r}} is n!r!(n−r)!\dfrac{n!}{r!\left( n-r \right)!} . The combination is used to find the number of arrangements which can take place. Do remember the term n!n! means the product of all the natural numbers from 11 to nn. Mathematically it is written as: n!=1×2×3.......×(n−1)×(n)n!=1\times 2\times 3.......\times \left( n-1 \right)\times \left( n \right)