math:combination & Permutation

小芝芝

2013-02-15 10:40 pm

Please to slove the about the math questions

1. How many different 3-letter strings can be formed from the letters of the
word 'QUIET' if no repetition of letters is allowed?

2. How many 4-digit numbers with 3 as the thousands digit can be cormed
from the digits 1,2,3,6,7 and 9 without repetition?

3.Among the letters of the word 'COMBINE' , 4 letters are taken and arranged
in a row. Find the number of possible arrangements if
(a) the letter B cannot be included,
(b) the letters B and C cannot be included at the same time

4. An examination paper consists of sections A, B and C. It is required to
answer 5 out of 7 questions in section A, 4 out 6 questions in section B,
3 out of 5 questions in section C. In how many ways can a candidate
attempt the paper if
(a) the candidate should answer all the sections?
(b) the candidate can choose 2 out of 3 sections to answer?

回答 (1)

賣女孩的火柴

2013-02-16 5:53 am

✔ 最佳答案

1.
Number of 3-letter strings
= 5P3
= 5!/2!
= 60

2.
Number of 4-digit numbers "3xyz"
= 1P1 x 5P3
= 1 x (5!/2!)
= 60

3.
(a)
Number of arrangements if "B" is not included
= 6P4
= 6!/2!
= 360

(b)
Number of arrangements if "C" is not included
= 6P4
= 6!/2!
= 360

Number of arrangements if both "B" and "C" are not included
= 5P4
= 5!/1!
= 120

The required number of arrangements
= 360 + 360 - 120
= 600

Alternative method :
Number of arrangements if there is no restriction
= 7P4
= 7!/3!
= 840

Number of arrangements if both "B" and "C" are included
= 5C2 x 4P4
= (5!/3!2!) x 4!
= 240

The required number of arrangements
= 840 - 240
= 600

4.
(a)
Number of ways
= 7C5 x 6C4 x 5C3
= (7!/5!2!) x (6!/4!2!) x (5!/3!2!)
= 21 x 15 x 10
= 3150

(b)
(Sections A and B / Sections B and C / Sections A and C)
Number of ways
= 7C5 x 6C4 + 6C4x 5C3 + 7C5 x 5C3
= 21 x 15 + 15 x 10 + 21 x 10
= 315 + 150 + 210
= 675

參考: 賣女孩的火柴

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