nCr and nPr 2
2011-01-03 6:21 am
(Q5) There are 4 boys and 4 girls. If 6 ofthem are choosen to line up, are the numbers of permutations in the followingtwo cases equal? Explain briefly. Case 1: The first two people in the line areboys, and the rest are girls. Case 2: Two boys stand at the two ends of theline, and the rest are girls. (Ans: Yes) (Q6) A zoo imported 7 kinds of animals:lion, tiger, elephant, giraffe, zebra, hippopotamus and chimpanzee. They willbe put into cages arranged in one row. If the elephant and the giraffe must beput in the cages at the two ends of the row, also the zebra and the tigercannot be put in the cages next to each other, how many ways are there to putthe animals in the cages? (Ans: 144) (Q7) 4 letters are selected from the wordCONFIRM. Find the number of combinations if vowel(s) should be selected. (Ans:30) (Q8) The following shows the agedistribution of the members in a gymnastics centre. 6 of them are selected toform a team. Find the number of combinations if at most two members aged 19 areselected. 17 age: 3 people / 18 age: 4 people / 19age: 6 people(Ans: 658)
回答 (1)
2011-01-03 8:24 am
✔ 最佳答案
(Q5)Case 1:
To arrange 2 boys in the first two places.
No. of permutations = 4P2 = 4!/(4 - 2)! = 12
To arrange 4 girls in the 4 last places,
no. of permutations = 4P4 = 4! = 24
Hence, no. of permutations of case 1 = 12 x 24 = 288
Case 2:
To arrange 2 boys in the two ends,
no. of permutations = 4P2 = 4!/(4 - 2)! = 12
To arrange 4 girls in the 4 middle places,
no. of permutations = 4P4 = 4! = 24
Hence, no. of permutations of case 2 = 12 x 24 = 288
Hence, the numbers of permutations in the two case are both equal to 288.
=====
(Q6)
To put elephant and giraffe at the two ends of the row.
No. of ways = 2P2 = 2! = 2
To arrange the rest 5 animals randomly into 5 cages,
no. of ways = 5P5 = 5! = 120
To arrange the rest 5 animals into 5 cages with zebra and tiger next to each other, it is considered as to combine zebra and tiger as one unit and arrange the 4 units with the internal permutation of the zebra and tiger.
No. of ways = 4P4 x 2P2 = 4! X 2! = 48
To arrange the rest 5 animals into 5 cages with zebra and tiger not next to each other,
no of ways = 120 - 48 = 72
Total number of ways to put the 7 animals to the cage
= 2 x 72
= 144
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(Q7)
Among the 7 letters, there are 2 vowels.
No. of combinations to select 4 letters randomly
= 7C4
= 7!/(7 - 4)!3!
= 35
No. of combinations to select 4 letters without vowel
= 5C4
= 5!/(5 - 4)!4!
= 5
No. of combinations to select 4 letter with vowel(s)
= 35 - 5
= 30
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(Q8)
If no aged 19 mmber is selected, 6 people are selected from the 7 aged 17-18 people.
No. of combinations = 7C6 = 7
If 1 aged 19 member is selected from the 6 aged 19 people, 5 people are selected from the 7 aged 17-18 people.
No. of combinations = 6C1 x 7C5 = 6 x 21 = 126
If 2 aged 19 member is selected from the 6 aged 19 people, 4 people are selected from the 7 aged 17-18 people.
No. of combinations = 6C2 x 7C4 = 15 x 35 = 525
Total number of combinations
= 7 + 126 + 525
= 658
參考: micatkie
收錄日期: 2021-04-13 17:44:38
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