Permutation

2011-01-16 11:30 pm
The question and solution are as follows:


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I don't know why it's 5*4*3*1 in case 1 and 4*4*3*2 in case 2? Please explain. Also, do you have any alternative way to solve it? thx!

回答 (1)

2011-01-17 12:05 am
✔ 最佳答案
form 3 students...question is:
从0,1,2,5,6,9十个数字中任意选择4个数字(每个数字最多重复1次),其等于的结果数字是四位偶數的组合分别有多少?
sol
for case 1: 选0
0不能在多位数中排首位,但0可以放在百位,十位&個位
而N1只有0,2和6是偶數
四位偶數的组合有
=5*4*3*1=60
for case 2:不选0
因0不能在多位数中排首位
同上,N1只有0,2和6是偶數 四位偶數的组合有
=4*4*3*2=96
一共可以组成没有重复数字的四位偶數
=96+60
=156

2011-01-16 16:09:22 補充:
if not clear...
let n1=0 from case 1
n4=1/2/5/6/9
n3(from 1XXX)=2/5/6/9
n2(from 12XX)=5/6/9
n1=0
so 5*4*3*1=60
or
let n1=2 from case 2
because 0 is not putting in the four-digit num(first num.)
n4=1/5/6/9
n3(from 1XXX)=0/5/6/9
n2(from 10XX)=5/6/9
n1=2
so 4*4*3*2=96

2011-01-16 16:12:02 補充:
sor...
from case 2
n1(from 105X)=2/6
so 4*4*3*2=96


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