Combination and permutation

2009-08-31 6:04 am
(a) How many different numbers of 4 digits may be formed with the digits 0,1,2,5,6,7,8 if no digit is used more than once in any number.

(b) How many of the numbers formed in (a) are even.

(c) Find the sum of all the numbers formed in (a).
更新1:

(b) 420 A detailed answer for (c) is expected.

回答 (2)

2009-08-31 7:53 am
✔ 最佳答案
First digit 6 choices : all except 0
Second digit 6 choices : all except first digit chosen
Third digit 5 choices : all except first 2 digits chosen
Last digit 4 choices : all except first 3 digits chosen
Together (6)(6)(5)(4) = 720


Consider last digit first:
Case (I) last digit is zero
Last digit only 1 choice : 0
First digit 6 choices : all except zero
Second digit 5 choices : all except 0 and first digit chosen
Third digit 4 choices : all except the 3 digits chosen
Together (1)(6)(5)(4) = 120
Case (I) last digit is non-zero
Last digit only 3 choices : 2, 6, 8
First digit 5 choices : all except zero and last digit chosen
Second digit 5 choices : all except first and last digit chosen
Third digit 4 choices : all except the 3 digits chosen
Together (3)(5)(5)(4) = 300
Case (I) + Case (II) = 420


Consider the case when zero is allowed in the first digit, there will be (7)(6)(5)(4) ways = 840 to form the numbers. Since there are 7 numbers to choose from, each number will appear 840/7 = 120 times in each of the digit positions.
Now since zero is not allowed in the first digit, 120 of the 0xxx cases were removed. Therefore each of the 6 numbers will appear 120/6 = 20 times less in the 2nd, 3rd and last digits. The number of occurrences in the first digit remains the same as only 0 is disallowed. So the total is calculated as:
120(1 + 2 + 5 + 6 + 7 + 8)(1000) + 100(1 + 2 + 5 + 6 + 7 + 8)(100) + 100((1 + 2 + 5 + 6 + 7 + 8)(10) + 100(1 + 2 + 5 + 6 + 7 + 8)
= 120000(29) + 10000(29) + 1000(29) + 100(29)
= 3801900

2009-08-31 6:51 am


(a) How many different numbers of 4 digitsmay be formed with the digits
0,1,2,5,6,7,8 if no digit is used more than once in any number.
Sol
6*6*5*4=720
or 7*6*5*4-1*6*5*4=840-120=720

(b) How many of the numbers formed in (a) are even.
(1) fourth is=0
7*6*5*1=210
(2) fourth is=2 or 6 or 8
5*5*4*3=300
210+300=510

(c) Find the sum of all the numbers formed in (a)
1+2+5+6+7+8=29
840/7=120,120/6=20
29*120*1111-29*20*111
=3801900




收錄日期: 2021-04-23 23:23:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20090830000051KK02193

檢視 Wayback Machine 備份