问题33.如果数字可以重复，那么数字1、2、3、4可以形成多少个不超过4321的四位数自然数?
解决方案：
Total numbers formed by these digits = 4 x 4 x 4 x 4 = 256 (At all 4 positions, 4 options)
Numbers exceeding 4321 are
At first position 4 is fixed
Case 1: 3 at second position
3 at 3rd digit = 4 numbers
4 at 3rd digit = 4 numbers
2 at 3rd and more than digit 1 at 4th = 3 numbers possible
Case 2: 4 at second position = 4 x 4 = 16 numbers (since 4 possibilities for 3rd as well as the 4th digits)
Such numbers = Case 1 + Case 2 = 4 + 4 + 3 + 16 = 27
Total required such numbers = 256 – 27 = 229
问题34.如果不重复数字，数字0、1、3、5、7和9可以形成多少个六位数?其中有多少可以被10整除?
解决方案：
First digit = 5 options (except 0)
Remaining = 5! (since 5 positions, 5 digits)
Total numbers = 5 x 5! = 600
Divisible by 10 = 0 at 6th position, remaining 5 digits and 5 positions – 5! = 120 numbers
问题35.如果在六个面上分别标记三个1至6的三个六个面对的骰子，则求出可能结果的总数。
解决方案：
For each dice, number of possible outcomes = 6
Three dice = 6 x 6 x 6 = 216 possibilities
问题36.投掷硬币3次，并记录结果。有多少可能的结果?如果抛硬币四次，可能有多少结果?五次? n次?
解决方案：
At each toss, 2 different possible outcomes
3 times coin tossed = 2 x 2 x 2 = 23 = 8
4 times = 2 x 2 x 2 x 2 = 24 = 16
5 times = 2 x 2 x 2 x 2 x 2 = 25 = 32
n times = 2n
问题37.如果可以用相同的数字重复数字，数字1,2,3,4,5可以形成多少个四位数?
解决方案：
For each digit = 5 possible numbers
So, since 4 digits = 5 x 5 x 5 x 5 = 625 numbers
问题38.使用数字0、1、3、5、7可以形成多少个三位数数字，而每个数字可以重复多次?
解决方案：
For 1st digit = 4 possible
For 2nd & 3rd = 5 possible
So, 4 x 5 x 5 = 100 such numbers
问题39.当一个数字可以重复任意多次时，可以从数字0、1、2、3、4、5形成多少个小于1000的自然数?
解决方案：
Given: Total number = 6
As we know that the natural ten’s than 1000 can be 1, 2 and 3 digit numbers
So, 0 cannot be the first digit of the 3-digit number.
Now, the hundred’s place can be filled with any of the 5 digits = 5 ways
The ten’s place can be filled with any of the 6 digits = 6 ways
The unit place can be filled with any of the 6 digits = 6 ways
Therefore, the total 3 digit number = 5 x 6 x 6 = 180
The total 2 digit number = 5 x 6 = 30
The total 1 digit number = 5
So, 180 + 30 +1 = 215 natural numbers.
问题40.使用数字0到9可以构造多少个五位数的电话号码。如果每个数字都以67开头并且没有数字出现一次以上?
解决方案：
Need to select digits for 3rd, 4th and 5th positions
So, 8 options for 3rd position and 7 for 4th position and 6 for 5th one
Total numbers are = 8 x 7 x 6 = 336
问题41：找出在5个孩子中分配8种不同玩具的方式数量。
解决方案：
Each toy has 5 options.
So, 8 toys = 58 = 390625
问题42.查找在7个信箱中可以张贴5个字母的方式的数量。
解决方案：
Total number of letters = 5
Total number of letter box = 7
So, each letter has 7 options and 5 total letters = 75 = 16807
问题43：掷三个骰子。找到至少一个骰子显示5的可能结果数。
解决方案：
Total number of outcomes = 6 x 6 x 6 = 216
Outcomes having no 5 = 5 x 5 x 5 = 125
Required outcomes = 216 – 125 = 91
问题44：找出将20个球放入5个盒子中的总方式，以便第一个盒子只包含一个球。
解决方案：
First box = 20 options
Other 19 balls have = 4 options = 419 arrangements for 4 boxes
Total ways = 20 x 419
问题45.在三个盒子中可以用多少种方式分配5个不同的球?
解决方案：
Total number of balls = 5
Total number of boxes =3
So, each ball has 3 options for the boxes = 35 = 243
问题46.在4个信箱中可以以几种方式张贴7个字母?
解决方案：
Total number of letters = 7
Total number of letter box = 4
So, each letter has 4 options and 7 total letters = 47 = 16384
问题47.何时可以在5个学生中以4种方式分配4个奖项?
(i)没有学生获得不止一个奖项吗?
(ii)一个学生可以获得多少个奖品?
(iii)没有学生获得所有奖品吗?
解决方案：
(i) So, 1 student does not get a prize
Choose that student among 5 students 5 different possible ways
Distribute among remaining students = 4! ways = 24
Total ways = 5 x 24 = 120
(ii) Every prize has 5 options
4 prizes = 5 x 5 x 5 x 5 = 625
(iii) Subtract the case of a student to get all prizes from (ii)
Selecting a student to give all prize = 5 ways
= 625 – 5 = 620
问题48.大厅里有10盏灯。它们中的每一个都可以独立打开。找到可以照亮大厅的方式。
解决方案：
Each lamp has 2 possibilities = on/off
Total ways = 210 = 1024
One way, all off = need to subtract it
1024 – 1 = 1023 ways