permuataion/combination

2010-04-23 9:44 am

(1)Five boys and three girlsa are about to form a queue. If the boys must stand next to one another, how many ways are there to form such a queue?

ans: 2880

(2)Five people a,b,c,d,e are going to line up in a queue. If a and b must be at the two ends respectively, in how many different ways can such a queue be formed?
A.72 B.48 C.36 D.20 E.12

(3)Find the total number of words formed by abracadabra taken all at a time. In how many of these will no two consonants appear together?

ans: 83160,180

回答 (1)

用戶2053

2010-04-23 11:03 am

✔ 最佳答案

1) (5 boys) have 5! = 120 ways
(5 boys) , (girl) , (girl) , (girl) have 4! = 24 ways
Total 120 x 24 = 2880 ways

2) (c d e) have 3! = 6 ways ,
a (c d e) b or b (c d e) a , 2 ways ,
Total 6 x 2 = 12 ways (E)

3) For 'abracadabra' , there are 11 letters contain 5 'a' , 2'b' , 1 'c' , 1 'd' & 2 'r' ,
The total number of words formed by abracadabra taken all at a time
= 11! / [(5!)(2!)(2!)] = 83160
b c d r are consonants , there are 6 consonants , 5 'a' ,
so all 'a' are between two consonants , for example b a b a x a d a r a r,
Total 6! / [(2!)(2!)] = 180 ways

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