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標題: [Math & Stat]Permutations [打印本頁]

作者: 雞 時間: 10/9/2008 07:30 PM
標題: [Math & Stat]Permutations
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? (720)
b) How many of the number formed in (a) are even? (420)
c) Find the sum of all the numbers formed in (a).

請指教(c), thanks

作者: Requiem 時間: 10/9/2008 07:58 PM
(1+2+5+6+7+8)x1000x(6x5x4)+(1+2+5+6+7+8)x100x(5x5x4)+(1+2+5+6+7+8)x10x(5x5x4)+(1+2+5+6+7+8)x1x(5x5x4)
=3801900

來亂

作者: 雞 時間: 10/9/2008 08:11 PM

原帖由 Requiem 於 10/9/2008 19:58 發表

 登錄/註冊後可看大圖

(1+2+5+6+7+8)x1000x(6x5x4)+(1+2+5+6+7+8)x100x(5x5x4)+(1+2+5+6+7+8)x10x(5x5x4)+(1+2+5+6+7+8)x1x(5x5x4)
=3801900

來亂

有沒有step..
看不明 =.="

作者: sapphire 時間: 10/9/2008 09:05 PM
as 0 cannot be the thousand digit, there is 6 ways to choose the thousand digit, and 6 * 5 * 4 ways to choose the other digits.

sum of thousand digit of all the numbers = (1 + 2 + 5 + 6 + 7 + 8) * 6 * 5 * 4 * 1000

since 0 has no value, and one of the number is already taken by the thousand digit, there is only 5 numbers left which will have values for hundred digits (the position of 0 in other digit does not affect the results of this sum)

sum of hundred digit of all the numbers = (1 + 2 + 5 + 6 + 7 + 8) * 5 * 5 * 4 * 100

similarly for the other two digits

sum of ten digit of all the numbers = (1 + 2 + 5 + 6 + 7 + 8) * 5 * 5 * 4 * 10
sum of last-digit of all the numbers = (1 + 2 + 5 + 6 + 7 + 8) * 5 * 5 * 4
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due to the poor writing skill, hope that u can understand it

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