Assume that there are 6 numbers of 1-mark questions, 5 numbers of 2-mark questions and 5 numbers of 5-mark questions. How many combinations of 10-mark question papers can be formed with at least 3 numbers of 1-mark question and 1 number of 2- mark question?
Answers
Step-by-step explanation:
Consider 3 questions
ABC
If A=1mark
Then
9 marks can be distributed among B and C in 8 different ways, since we do not consider the possibility of zero marks. That is both B and C holds at-least 1 mark.
Hence for A=1marks we have 8 ways of distributing 10 marks among 3 questions.
Similarly
For A=2marks we will have 7 ways of distributing remaining 8 marks among B and C, Hence 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 and so on
:
:
Hence for A=9 marks we will have only one way of distributing 10 marks among 3 question A, B,C.
Thus the total number of ways of distributing 10 marks among 3 questions such that each has at-least 1 marks is
=8+7+6+5+...+1
= Sum of first 8 natural numbers.
=28(8+1)
=272
=36