數學疑難 [復制鏈接]

The prime ministers A,B,C,D,E,F and G of 7 countries will address at a summit meeting.
(a) Find the number of arrangements that can be made so that
(i) A will speak before C,
(ii) A will speak before C and C before E.
(b) In how many of those ways in (a) (ii) will C speak immediately after A?

引用第0樓Skyline於2006-07-15 16:44發表的“數學疑難”:
The prime ministers A,B,C,D,E,F and G of 7 countries will address at a summit meeting.
(a) Find the number of arrangements that can be made so that
(i) A will speak before C,
(ii) A will speak before C and C before E.
(b) In how many of those ways in (a) (ii) will C speak immediately after A?

A question on permutation

have u learnt sth like 5P2, 10C3，　４！　 ??

e.g. try to list the following table and solve for a(i)

１　２　３　４　５　６　７
ａ　　　　　　　　　　　　　　　　　５Ｐ０　Ｘ　６！　＝　７２０
　　ａ　　　　　　　　　　　　　　　５Ｐ１　Ｘ　５！　＝　６００
　　　　ａ　　　　　　　　　　　　　５Ｐ２　Ｘ　４！　＝　４８０
　　　　　　ａ　　　　　　　　　　　５Ｐ３　Ｘ　３！　＝　３６０
　　　　　　　　ａ　　　　　　　　　５Ｐ４　Ｘ　２！　＝　２４０
　　　　　　　　　　ａ　　　　　　　５Ｐ５　Ｘ　１！　＝　１２０　
　　　　　　　　　　　　ａ　　　　　Impossible

So total favoritable outcomes = 2520

note that possible outcomes = ７！=5040

as by common sense, either A is in front of C, or behind C. And the theorical probability is 1/2, so the answer makes sense.

others u try to solve by urself first, if still donno, just ask.

明解了, thz

Total=7P7=5040

A講第nth可能=5040/7=720

１　２　３　４　５　６　７
ａ　                             =>720*6/6 = 720　　　　　
  　ａ　　                     =>720*5/6 = 600 (A講第2即C在六個可能中有5個可能講後)　　　　　
　 　     ａ　　　　　　　　=>720*4/6 = 480　　　　　
　　　　　　ａ　　　　　　=>720*3/6 = 360　　　　　
　　　　　　　　ａ　　　　=>720*2/6 = 240　　　　　
　　　　　　　　　　ａ　　=>720*1/6 = 120　　　　　　
　　　　　　　　　　　　ａ=>720*0/6 = 0　　　　
即 720+600+480+360+240+120+0 = 2520 (out of 5040 possible choice)　

similar solution as kylau for reference

(a)(ii)
我咁計:
4! x 5 + 4! x 4 x 2 + 4! x 3 x 3 + 4! x 4 x 2 + 4! x 5 =840

仲有冇其他計法

引用第4樓Skyline於2006-07-15 20:58發表的“”:
(a)(ii)
我咁計:
4! x 5 + 4! x 4 x 2 + 4! x 3 x 3 + 4! x 4 x 2 + 4! x 5 =840
仲有冇其他計法

Do you get the ans?
I just count the possible choice and the number is much less than your ans.

===================
Oh, I miss sth...so I just only count the small no.
I get the ans now, ...
Share with you, although not too good method
still fix a

１　２　３　４　５　６　７
ａ　                                 5+4+3+2+1=15#
　ａ　　               　　　   4+3+2+1=10
　 　   ａ　　　　　　　　     3+2+1=6
　　　　　　ａ　　　　　　   2+1=3
　　　　　　　　ａ　　　　　1 (only one solution)
　　　　　　　　　　ａ　　   0 (no solution)
　　　　　　　　　　　　ａ　0 (no solution)　
#5 means put C at "2" and there will be 3,4,5,6,7 (5 choice) for E
4 means put C at "3" and there will be 4,5,6,7 (4 choice) for E
other are similar
the total is 35, with consider other speckers can speck in any order, 4P4
=> 35*4P4 = 840

ur method is ok

and this is the best method i can think of

i think skyline's idea is that
now instead of fixing A first, fix C

１　２　３　４　５　６　７
ｃ　                   　　　　　impossible
　ｃ　　               　　　　　１ｘ５ｘ４！
　 　   ｃ　　　　　　　　　　　　２ｘ４ｘ４！　
　　　　　　ｃ　　　　　　　　　　３ｘ３ｘ４！
　　　　　　　　ｃ　　　　　　　　４ｘ２ｘ４！　
　　　　　　　　　　ｃ　　　　　　５ｘ１ｘ４！　
　　　　　　　　　　　　ｃ　　　　impossible

say for C is on the 2nd,
the 1 means there's a possible case for A to be in front of C
the 5 means there's 5 possible cases for E to be behind C
the 4! means the arrangement of the other 4 letters

引用第5樓ballballfull於2006-07-15 21:28發表的“”:
Do you get the ans?
I just count the possible choice and the number is much less than your ans.
.......

=3=

無聊一問：

ballballfull你係讀maths/stat major的？

engine subject

asl app. math
al p. math

math is fun

If I can study more, I think stat. is a good choice.
However, I forget most knowledge of AL math.

咁深既........

完全唔知咩黎@@"

引用第9樓ballballfull於2006-07-16 21:12發表的“”:
engine subject
asl app. math
al p. math
.......

AL maths 我認為只有differentiation, integration 同Probability有用
其他唔記得都無咩所謂

matrix 個d
讀engineering subject 有用
cad/cam