Preputation , 有答案, 只想要解釋

2007-01-08 5:42 am

3.How many different integers can be formed by using 0, 1, 2, 3, 4, and 5 (not allow repetition) if the integer is:

a) 5 digits (i.e. 43215, 43012…)?

5x 5P4 = 600

b) 5 digits odd number?

4x 4P3 x 3 = 288

想問點解 ?

回答 (1)

myisland8132

2007-01-08 6:30 am

✔ 最佳答案

(a)
The first place cannot place the digit 0
(for example 01234 does not accept)
so, we have 5 choice
At the second place, we have 5 choice
the third place, 4 choice
the fourth place, 3 choice
the fifth place, 2 choice
Total different integers can be formed
=5*5*4*3*2
=5*5P4
=600
(b)
For 5 digits odd number
the fifth digit should be odd, we have three choice (1,3,5)
let say the fifth place is 1
then
At the first place, we have 4 choice
At the second place, we have 4 choice
the third place, 3 choice
the fourth place, 2 choice
So the different integers can be formed that the last digit is 1
=4*4*3*2
=4*4P3
the same argument can be used for last digit 3 and 5
so
the total different integers can be formed
=4x 4P3 x 3
=288

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