Maths S1 M/J10 Solve please

[zwitterion7]

Can someone please solve Q7 part 4 and 5.
http://www.xtremepapers.me/CIE/Internat ... _qp_62.pdf

 

[remz]

Part 4 of Question 7
As we have already obtained 9P3 = 504 ways in the previous part, in this part they are telling us that one of the 3 cards is Pink. So we imagine that we have ( P , C1 , C2 ) - Remember all of them are of different color. As we know that pink is already in the arrangement, so its 8C2. Next we take 3! due to its arrangement. So 3! x 8C2 is 168.

I am trying to solve part 5 of the question but that seems like a tuff nut to crack!

 

[ahmed t]

for part 5
all u do is find the number of ways they are together thenn (iii)-ans
so if blue and pink are together then u must choose 1 other card from the 7
so
7C1*2*2
since you can arrange the cards in 2*2 ways
so you will get 28
so
504-28=476

 

[mickysharif]

God I have no idea what "9P3" is, I don't think I have ever even done that on a calculator yet.. can someone please explain the difference between 9P3 and 9C3, because It think im going to do really really poorly in the exam 2morrow.

 

[remz]

God I have no idea what "9P3" is, I don't think I have ever even done that on a calculator yet.. can someone please explain the difference between 9P3 and 9C3, because It think im going to do really really poorly in the exam 2morrow.

P is permutation where arrangement is important (ordering is important). Usually found when you are placing people in a row or whichever situation requires order
However combination (C) is where arrangement is NOT important and usual events involving it are people being selected in a group, etc.