1. Under the following conditions, how many 3-digit numbers can be formed by using 4 numerals 2,4,6 and 8? The multiplication rule in the counting principle should be used.
a) Each numeral can only be used once.
b) Each numeral can be used more than once.
2. By using 10 cards marked 0 to 9 respectively.
a) how many 4-digit numbers that are divisible by 10 can be formed?
b) how many 5-digit numbers that are not divisible by 10 can be formed?
Permutation and Combination
2010-10-16 6:44 pm
回答 (1)
2010-10-17 2:29 am
✔ 最佳答案
1a) 4P3 = 4!/(4-3)!=241b) 4^3=64
2a) if divisible by 10, the last digit must be 0
choose 3 other digits from 9 cards (0 is choosen for the last digit)
9P3=9!/(9-3)!
=504
2b) Total no. of ways = 10P5 =10!/(10-5)!=30240
no. of ways divisible by 10= 9P4 (similar to part a)=9!/(9-4)!=3024
(Total no. of ways - no. of ways divisible by 10)
=30240-3024 =27216 //
收錄日期: 2021-04-23 22:11:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101016000051KK00383