Thanks so much i really struggle with this topic.Amongst the first eight positive integers, {1, 2, 3, ..., 8}, there are four odd integers {1, 3, 5, 7} and four even integers {2, 4, 6, 8}. The selection of two odd integers and two even integers can only be achieved by selecting two integers from each group. Order is not important. There are therefore
ways in which this can be done.
thank you very much, i suck at this topic.Amongst the first eight positive integers, {1, 2, 3, ..., 8}, there are four odd integers {1, 3, 5, 7} and four even integers {2, 4, 6, 8}. The selection of two odd integers and two even integers can only be achieved by selecting two integers from each group. Order is not important. There are therefore
ways in which this can be done.
No worries, yashbb. Bear in mind that the logic / reasoning in this topic is quite different from that in most other parts of HSC maths, so you are far from alone in finding perms and combs difficult.Thanks so much i really struggle with this topic.
I agree with what CM_Tutor mentioned above about this being one of the hardest topics in HSC maths. What I used to do was break down the question into smaller parts. Yes for this question, it might not be necessary, but for longer questions, especially where there's more involved, it will definitely help rather than looking at one big chunk of information and stressing.Four numbers are to be selected from the set of the first eight positive integers. Find how many possible combinations there are if:
there are two odd numbers and two even numbers
Thank you! Really appreciate it!I agree with what CM_Tutor mentioned above about this being one of the hardest topics in HSC maths. What I used to do was break down the question into smaller parts. Yes for this question, it might not be necessary, but for longer questions, especially where there's more involved, it will definitely help rather than looking at one big chunk of information and stressing.
For example, in this question, you may want to draw 2 collections of odd and even numbers, so 1 3 5 7 2 4 6 8
Then you could take each case on its own and consider how many ways you can choose 2 from each, so 4C2 from the odds and 4C2 from the evens; then, you would multiply them.