Help me!
Please to slove the following math questions
1. In how many ways can 2 boys and 3 girls be seated in a now such that
children of the same sex must not sit next to each other?
2. 5 couples are sitting in a now. If males and females must sit alternatively,
find the number of possible arrangements.
3.All the letters of the word 'PLACES' are rearranged . Find the number of
possible arrangements if
(a) the first letter is a vowel
(b) the letters A and E are side by side
4.All the letters of the word 'EQUATIONS' are rearranged. Find the number of
possible arrangements if
(a) the letter in the middle must be a vowel
(b) both ends are vowels
5. 10 models are liming up in a row on a stage. In how many ways can
they be arranged if
(a) two particlar models must stand next to each other?
(b) two particular models must be separated?
Math EX.
2013-02-15 10:19 pm
回答 (2)
2013-02-16 6:44 am
✔ 最佳答案
1.(1) Arrange the 3 girls (G) as : G_G_G (3P3).
(2) Arrange the 2 boys to the two "_" (2P2).
The required number of ways
= 3P3 x 2P2
= 3! x 2!
= 12
2.
(1) Arrange the 5 males (M) as : M_M_M_M_M (5P5).
(2) Put any 4 females (F)into the 4 "_" (5P4).
(3) put the last female as the first person or the last person (2P1).
The required number of ways
= 5P5 x 5P4 x 2P1
= 5! x (5!/1!) x (2!/1!)
= 120 x 120 x 2
= 28800
Alternative method :
MFMFMFMFMF or FMFMFMFMFM
The required number of ways
= 5P5 x 5P5 + 5P5x 5P5
= 5! x 5! + 5! x 5!
= 120 x 120 + 120 x 120
= 28800
3.
(a)
(1) Put one of the two vowels ('A' and 'E') as the first letter (2P1).
(2) Arrange the rest 5 letters (5P5).
The required number of arrangements
= 2P1 x 5P5
= (2!/1!) x 5!
= 2 x 120
= 240
(b)
(1) Put A and E as a pair with internal permutation = 2P2
(2) Arrange the pair and the rest 4 letter (5P5).
The required number of arrangements
= 2P2 x 5P5
= 2! x 5!
= 2 x 120
= 240
4.
(a)
Out of the 9 letters, there are 4 vowels : U, A, I and O.
(1) Put 1 of the 4 vowelsat the middle (4P1).
(2) Arrange the rest 8 letters (8P8).
The required number of arrangements
= 4P1 x 8P8
= (4!/3!) x 8!
= 4 x 40320
= 161280
(b)
(1) Put 2 of the 4 vowels at both ends (4P2).
(2) Arrange the rest 7 letters (7P7).
The required number of arrangements
= 4P2 x 7P7
= (4!/2!) x 7!
= 12 x 5040
= 60480
5.
(a)
(1) Put the two particular model as a pair with internal permutation = 2P2
(2) Line up the pair and the rest 8 model (9P9).
The required number of arrangements
= 2P2 x 9P9
= 2! x 9!
= 2 x 362880
= 725760
(b)
(1) Arrange the rest 8 models (M) as _M_M_M_M_M_M_M_M_ (8P8).
(2) Put the 2 particular models in any 2 "_" out of 9 "_" (9P2).
The required number of arrangements
= 8P8 x 9P2
= 8! x (9!/7!)
= 40320 x 72
= 2903040
Alternative method :
The required number of arrangements
= (Number of arrangement without restriction) - (Number of arrangements in (a))
= 10! - 725760
= 3628800 - 725760
= 2903040
參考: 賣女孩的火柴
2013-02-16 2:21 am
1) 5! - 2!*3! = 108
2) 你意思佢地分開坐?
如果系: 5!*5! = 14400
3a) 5!*2 = 240
3b) 6! - 2!*5! = 480
4a) 5*8! = 201600
4b) 5C2*2* 7! = 100800
5a) 2!*9! = 725760
5b) 10! - 2!*9! = 2903040
2) 你意思佢地分開坐?
如果系: 5!*5! = 14400
3a) 5!*2 = 240
3b) 6! - 2!*5! = 480
4a) 5*8! = 201600
4b) 5C2*2* 7! = 100800
5a) 2!*9! = 725760
5b) 10! - 2!*9! = 2903040
參考: self
收錄日期: 2021-04-13 19:17:35
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