用数字0、1、2、3、4、5一共可以组成多少个没有重复数字且能被5整除的四位数

2025-03-13 18:07:00
推荐回答(5个)
回答1:

两种情况:
1)个位为0。从1,2,3,4,5中任取3个排在高三位,共有A(1,)*A(5,3)=1*5*4*3=60个。
2)个位为5。从1,2,3,4中取一个排在最高位,再从剩下的4个中取2个排在中间两位,
共有 A(1,1)*A(4,1)*A(4,2)=1*4*4*3=48个。

所以,所求的四位数共有 60+48=108 个。

回答2:

P(5,3)+P(4,1)*P(4,2)
=60+4*12
=108
用数字0、1、2、3、4、5一共可以组成108个没有重复数字且能被5整除的四位数

回答3:

个位上是0的有5×4×3=60个
个位上是5的数有5×5×4=100个
一共有60+100=160个

回答4:

360个没有重复的4位数,其中120个可以被5整除,分别是:
0125
0135
0145
0215
0235
0245
0315
0325
0345
0415
0425
0435
1025
1035
1045
1205
1230
1235
1240
1245
1250
1305
1320
1325
1340
1345
1350
1405
1420
1425
1430
1435
1450
1520
1530
1540
2015
2035
2045
2105
2130
2135
2140
2145
2150
2305
2310
2315
2340
2345
2350
2405
2410
2415
2430
2435
2450
2510
2530
2540
3015
3025
3045
3105
3120
3125
3140
3145
3150
3205
3210
3215
3240
3245
3250
3405
3410
3415
3420
3425
3450
3510
3520
3540
4015
4025
4035
4105
4120
4125
4130
4135
4150
4205
4210
4215
4230
4235
4250
4305
4310
4315
4320
4325
4350
4510
4520
4530
5120
5130
5140
5210
5230
5240
5310
5320
5340
5410
5420
5430

回答5:

108
5*4*3+5*4*3-4*3=180

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