hi tys question but im facing abit of difficulty
qn: group of diplomats chosen to represent 3 islands, K,L and M . group is to consist of 8 diplomats and is chosen from a set of 12 diplomats consisting of 3 from k, 4 from L and 5 from M . Find the number of ways the group can be chosen if it includes
a) diplomats from L and M only
i did it like this:
case 1: 4 from M, 4 from L = 5c4 X 4c4 = 5
case 2: 5 from M , 3 from L = 5c1 X 4c3 = 20
therefore total ways is 25 .
but ans is 9 ( from tys)
b)at least 4 diplomats from M
i did it like this
C1: 4M, 4 L =5
C2: 4M, 3L, 1K = 5c4 . 4c3 . 3c1 =5.4.3=60
C3: 4M, 3K, 1L = 5c4 . 3c3 . 4c1=5.1.4=20
C4: 4M,2L,2K = 5c4 . 4c2. 3c2 =5.6.3=90
C5:4M,2K,2L=90
C6:5M,3L = 5c5. 4c3= 1.4=4
C7: 5M,3K = 5c5 . 3c3 =1
C8: 5M,1K,2L = 5c5 . 3c1 . 4c2 =1.3.6=18
C9: 5M,2K,1L = 5c5 . 3c2 . 4c1= 1.3.4=12
total is 290 .
ans from tys is 210
am i wrong or is the tys wrong? if so , where have i gone wrong? teach me thanks xD
Hi,
This question comes from paper 2 of 2008 exam. It is a selection type of problem for this topic.
(a) We have a total of 9 diplomats from L and M, of which we choose 8 to form the group.
(b) We need to consider 2 cases:
1. 4 diplomats from M - choose 4 out of 5 from M, then choose 4 out of 7 from K & L combined;
2. 5 diplomats from M - simply choose 3 out of 7 from K & L combined.
An important pointer for solving P&C problems is to think as simply as possible by applying addition (whenever there are cases to consider), multiplication (whenever there are sequences of events to consider), and mutual exclusion principles.
Thanks.
Cheers,
Wen Shih
Hi,
You may wish to read my 4-part postings on P&C at:
http://www.sgforums.com/forums/2297/topics/356377?page=3
Thanks.
Cheers,
Wen Shih