问题17.有多少个三位数的数字?
解决方案：
Ways of selecting 100th place = 9P1 = 9 (Selecting from all digits except 0)
Ways of selecting 10th place = 10P1 = 10 (Selecting from all digits)
Ways of selecting unit pace = 10P1 = 10 (Selecting from all digits)
Total 3-digit numbers possible = 9 x 10 x 10 = 900
问题18.有多少个三位数的奇数?
解决方案：
Ways of selecting 100th place = 9P1 = 9 (Selecting from all digits except 0)
Ways of selecting 10th place = 10P1 = 10 (Selecting from all digits)
Ways of selecting unit pace = 5P1 = 5 (Selecting from all odd digits)
Total 3-digit odd numbers = 9 x 10 x 5 = 450
问题19.如果可以制作多少个不同的五位数字车牌
(i)第一位数字不能为零，并且不允许重复数字，
(ii)首位数字不能为零，但可以重复数字吗?
解决方案：
(i) First digit = 9P1 = 9 ways (all digits except zero)
Second digit = 9P1 = 9 ways (all digits except the first digit of the license plate)
Third digit = 8P1 = 8 ways (all digits except the first and second digits of the license plate)
Fourth digit = 7P1 = 7 ways (all digits except the first, second and third digits of the license plate)
Fifth digit = 6 ways (all digits except the first, second, third and fourth digits of the license plate)
Total ways = 9 x 9 x 8 x 7 x 6
= 27216
(ii) First digit = 9P1 = 9 ways (all digits except zero)
Second digit = 10P1 = 10 ways (all digits)
Third digit = 10P1 = 10 ways (all digits)
Fourth digit = 10P1 = 10 ways (all digits)
Fifth digit = 10P1 = 10 ways (all digits)
Total ways = 9 x 10 x 10 x 10 x 10
= 90,000
问题20.如果不允许重复数字，数字3、5、7、8、9可以形成大于7000的四个四位数数字吗?
解决方案：
First digit = 3P1 = 3 ways (only 7, 8, 9 are possible digits to make it greater than 7000)
Second digit = 4P1 = 4 ways (all 5 digits except the first digit of the selected number)
Third digit = 3P1 = 3 ways (all 5 digits except the first and second digits of the number)
Fourth digit = 2P1 = 2 ways (all 5 digits except the first, second and third digits of the number)
Total numbers = 3 x 4 x 3 x 2
= 72
问题21.如果不允许重复数字，可以用大于8000的数字3、5、7、8、9形成多少个四位数数字?
解决方案：
First digit = 2P1 = 2 ways (only 8, 9 are possible digits to make it greater than 8000)
Second digit = 4P1 = 4 ways (all 5 digits except the first digit of the selected number)
Third digit = 3P1 = 3 ways (all 5 digits except the first and second digits of the number)
Fourth digit = 2P1 = 2 ways (all 5 digits except the first, second and third digits of the number)
Total numbers = 2 x 4 x 3 x 2
= 48
问题22.六个人可以以几种方式连续坐下?
解决方案：
Let’s consider 6 seats in the given row
First seat = 6P1 = 6 options of people
Second seat= 5P1 = 5 options (the person already sitting on first seat can not sit here)
Third seat = 4P1 = 4 options (people sitting on first and second can’t sit on third)
Fourth seat = 3P1 = 3 options
Fifth seat = 2P1 = 2 options
Sixth seat = 1P1 = 1 option
Total ways = 6 x 5 x 4 x 3 x 2 x 1 = 6! = 720
问题23.可以形成多少个不同数字的9位数字?
解决方案：
1st digit = 9P1 = 9 possibilities (zero is not possible)
2nd digit = 9P1 = 9 possibilities (except first digit)
3rd digit = 8P1 = 8 possibilities (except first & second digits selected)
4th digit = 7P1 = 7 possibilities (except first, second, third digits selected)
for all subsequent i-th position possibilities keep on decreasing by 1 so,
eventually for 9th digit, 2 possibilities
Total numbers = 9 x 9 x 8 x ….. 2
= 9 x 9!
= 3265920
问题24.在不允许重复数字时，使用数字0、3、5、7可以形成小于1000的奇数个数字吗?
解决方案：
Case 1: 1 digit number
3P1 = 3 ways (since 0 is even so it can not be selected)
Case 2: 2 digit numbers
1st digit = 3P1 = 3 ways (except 0)
2nd digit = 2P1 = 2 ways (except first digit and 0)
= 3 x 2 = 6
Case 3: 3 digit numbers
1st digit = 3P1 = 3 ways (except 0)
2nd digit = 3P1 = 3 ways (except the first digit)
3rd digit = 2P1 = 2 ways (except the first and second digits)
4th digit = 1P1 = 1 ways (except the first, second, third digits)
Total numbers including odd and even = 3 x 3 x 2 x 1 = 18
Even such numbers included = No.s ending with zero
= Number of options for 3rd position is 1
= Number of options for 2nd position is 3
= Number of options for 1st position – 2
Even numbers = 1 x 3 x 2 = 6
= 18 – 6 = 12 (odd numbers)
Case 4: 4 digit numbers
No possibility (since they’ll be greater than 1000)
Total numbers = 3 + 6 + 12 + 0
= 21
问题25.有多少个3位数字，每个数字都不同，而每个数字都为奇数?
解决方案：
Odd digits are 1, 3, 5, 7, 9
So, the total number of odd digits = 5
Now, select 3 digits = 5P3 = 10 ways
Arrange these selected digits = 3! ways
Total numbers = 10 x 3! = 60
问题26.在不允许重复数字的情况下，数字4、5、6、7、8、9可以形成多少个不同的六位数?
解决方案：
First digit of six-digit number can be selected = 6 ways
Second digit of six-digit number can be selected = 5 ways
Third digit of six-digit number can be selected = 4 ways
Fourth digit of six-digit number can be selected = 3 ways
Fifth digit of six-digit number can be selected = 2 ways
And the last digit of six-digit number can be selected = 1 ways
So, the final arrangements of 6 digits = 6 x 5 x 4 x 3 x 2 x 1 = 720 ways
问题27.在不允许重复数字时，数字3、1、7、0、9、5可以形成多少个不同的六位数?
解决方案：
Possibilities for 1st digit = 5 (except 0)
Possibilities for 2nd digit = 5
Possibilities for 3rd digit = 4
Possibilities for 4th digit = 3
Possibilities for 5th digit = 2
Possibilities for last digit = 1
So, the total numbers = 5 x 5 x 4 x 3 x 2 x 1 = 600
问题28.在不允许重复数字时，可以用数字1,2,5,9,0形成多少个大于5000的四位数不同数字?
解决方案：
For 1st digit, ways = 2 ways (only 5 or 9)
For remaining places = 4P3 ways
To choose the digits = 4P3 x 3! = 24 arrangements
Total ways = 2 x 24 = 48
问题29.在工厂生产的物品的序列号应使用两个字母和四个数字(0到9)制成。如果字母要从六个英文字母中不取而代之，并且数字也不要在序列号中重复，那么可以有多少个序列号?
解决方案：
Arrangements for 2 letters = 2! x 6P2 = 30
Arrangements for 4 digits = 4! x 10P4 = 5040
Required numbers = 30 x 5040 = 151200
问题30.行李箱上的数字锁有3个轮子，每个轮子用10到0到10的数字标记。如果开锁是一个特定的三位数字序列，且没有重复，那么可能有多少个序列?另外，找到打开锁失败的尝试次数。
解决方案：
Ways of first number = 10
Ways of second number = 9 (except first)
Ways of third number = 8 (except first and second)
Total numbers = 10 x 9 x 8 = 720
Number of unsuccessful attempts = 719 (since only 1 will be correct possibility)
问题31.客户忘记了银行的自动柜员机(ATM)的四位数代码。但是，他记得该代码由数字3、5、6和9组成。找到获得正确代码所需的最大尝试次数。
解决方案：
Total number of digits = 4
So, the largest possible number of trials necessary to obtain the correct code = 4! = 24 attempts
问题32.如果一个人只被分配一份工作并且每个人都有能力完成一项工作，那么可以将三种工作I，II和III分配给三个人A，B和C?
解决方案：
Total number of jobs = 3
Total number of persons = 3
It is exactly same as arranging 3 objects, in 3 different positions = 3! ways = 6