Need help ... dunno how to describe the question :S (1 Viewer)

[boongsta]

Lauren is going to the UB (Universal Baccalaureat) Mathematics exam next week. The exam consists of 10 questions, four in section A, three in section B and three in section C. In how many orders can she answer the 10 questions if B)she does a section A question first, and another one last, c) she does all the section A questions together, all the section B questions together and all the section C questions together (but in any order naturally) d) She does one from each section as her first three questions.

i really need help with ths question ... not sure if u need to use the factorial function or the permutation one

 

[icycloud]

I'm a bit rusty on these problems, and my answers don't sound very right lol, so someone please check them for correctness!

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B)she does a section A question first, and another one last

There are 4 possibilities for Question 1, leaving 3 possibilities for Question 10, in between there are 8 possibilities for Question 2, 7 for Q3, 6 for Q4, etc...

Thus, the total number of possibilities is 4*3*8! = 483,840

c) she does all the section A questions together, all the section B questions together and all the section C questions together (but in any order naturally)

Number of ways section A questions can be done: 4P4
Number of ways section B questions can be done: 3P3
Number of ways section C questions can be done: 3P3

Ordering of Section A, Section B, Section C: 3P3

Therefore, total number of possibilities = 3P3 * (4P4 * 3P3 * 3P3) = 5,184

d) She does one from each section as her first three questions.

Number of ways to have the first three questions from each section:

4*3*3*6 = 216

Now consider the first three questions as one group,

So we now have 8 items, possibilities = 216 * 7P7 = 1,088,640

 

[boongsta]

thanks heaps for that it should work