Question

The number of different ways of distributing 10 marks among 3 questions, cash carrying at least 1 mark, is
a)72 b.) 71 c.) 36 d.) None.

Answer 1

Hint:- In this type of problem, we will check one-by-one for each and every mark awarded; as for 1 mark, it is equal to the 8 way; 2 marks , it is equal to (8-1) way and so on for the 9 marks.

Complete step-by-step answer:
Consider 3 question:-
     A, B, C If A = 1 mark;
Then, 9 marks can be distributed among B and C in 8 different ways, since we do not consider the possibility of zero marks. This is because both B and C hold at least 1 mark.
Hence for A= 1 marks, we have 8 ways of distributing 10 marks among 3 questions.
Similarly,
For A = 2 marks we will have 7 ways of distributing remaining 8 marks among B and S
Here there will be 7 ways of distributing 10 marks amongst 3 questions.
For A= 3, we will have 6 ways of distributing 10 marks among 3 questions.
For A = 4,we will have 5 ways.
For A = 5, we will have 4 ways.
For A = 6, we will have 3 ways.
For A = 7, we will have 2 ways.
For A = 8, we will have 1 way
For A = 9 marks, we will have only one way of distributing 10 marks among 3 questions A, B, c.
Thus the total, number of ways of distributing 10 marks among 3 questions such that each has at least 1 mark is
= 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1
= sum of first 8 natural numbers.
$ = \dfrac{{8\left( {8 + 1} \right)}}{2}$
$ = \dfrac{{8 \times 9}}{2} = 4 \times 9 = \boxed{36}$
$\left[ {\therefore {\text{sum}}\;{\text{of}}'n'{\text{natural}}\;{\text{number}}\; = \dfrac{{n\left( {n + 1} \right)}}{2}} \right]$

Note:- We know that, sum of the first n natural numbers that is 1+ 2+ ... + n = n(n+1) / 2, for n being a natural number.
So, sum of first ‘n’ natural number is: -\[\dfrac{n\left( n+1 \right)}{2}.\]
The number of ways of distributing 10 marks among 3 questions such that each has at least 1 mark can be calculated one-by-one for each and every mark awarded.