问题1.查找通过排列以下单词的所有字母形成的单词数：
(i)独立性
(ii)中级
(iii)安排
(iv)印度
(v)巴基斯坦
(vi)俄罗斯
(vii)系列
(viii)锻炼
(ix)君士坦丁堡
解决方案：
(i) Given: 12 letter word in which
N appear = 3 times
E appear = 4 times
D appear = 2 times
Remaining letters = once
Number of permutations of these letters = 12! / (3! 4! 2! ) = 1663200
(ii) Given: Total 12 letters there
I appear = twice
T appear =twice
E appear = thrice
Remaining = once
Number of permutations = 12! / (2! 2! 3!) = 19958400
(iii) Given: 7 letters in which
A appear = 2 times
R appear = 2 times
Remaining = once
Number of permutations = 7! / (2! 2!) = 1260
(iv) Given: 5 letters in which
I appear = twice
Remaining = once
Number of permutations = 5! / 2! = 60
(v) Given: 8 letters in which
A appear = 2 times
Remaining = once
Number of permutations = 8! / 2! = 20160
(vi) Given: 6 letters in which
S appear = twice
Remaining = once
Number of permutations = 6!/2! = 360
(vii) Given: 6 letters in which
S appear = twice
E appear = twice
Remaining = once
Number of permutations = 6! / (2! 2!) = 180
(viii) Given: 9 letters in which
E appear = 3 times
S appear = 2 times
Remaining = once
Number of permutations = 9! / (3! 2!) = 30240
(ix) Given: 14 letters in which
T appear = 2 times
O appear = 2 times
N appear = 3 times
Remaining appear = 1 time
Number of permutations = 14! / (2! 2! 3!) = 3632428800
问题2.在不改变元音和辅音的相对顺序的情况下，可以用多少种方式排列“代数”一词的字母?
解决方案：
Given: 3 vowels and 4 consonants present in the given word
Vowels (A, E, A) = 3!/2! = 3 ways of permutations
Consonants = 4! = 24 permutations
Permutations so that the given condition is satisfied = 3 x 24 = 72
问题3.“ UNIVERSITY”一词的字母可以组成多少个单词，元音保持在一起?
解决方案：
4 vowels present can be arranged in = 4!/2! ways
First vowel can have positions numbers 1 to 7 and
remaining will follow after that in consecutive order = 7C1 = 7 ways
Permutations of the consonants = 6!
Total words that satisfy given conditions = 6! x 7 x 4!/2! = 60480
问题4.在全长写的情况下，找到表达式a 3 b 2 c 4中字母排列的总数。
解决方案：
In a3 b2 c4,
a appear = 3 times
b appear = 2 times
c appear = 4 times
So, the total arrangements = 9! / (3! 2! 4!) = 1260
问题5.单词“ PARALLEL”的字母可以形成多少个单词，以使所有L不在一起?
解决方案：
Total permutations = 8! / (3! 2!) = 3360
Number of words in which all L’s together (Replace 3 L by one symbol
so this 1 symbol and 5 other letters left)
= 6! / 2! = 360
Desired words = 3360 – 360 = 3000
问题6.通过排列单词’MUMBAI’的字母可以形成多少个单词，以便所有M在一起?
解决方案：
Total number of letters present in the MUMBAI word = 6
Replace 2 M’s by 1 symbol = 5 letters left there
Number of words = 5! = 120
问题7.用数字1、2、3、4、3、2、1可以形成多少个数字，以便奇数位始终占据奇数位?
解决方案：
Given: 4 digits are odd and 4 odd positions
Arrange these 4 odd digits in 4 positions = 4!/(2! 2!) = 6
Arrange remaining 3 digits = 3!/2! = 3
Total permutations = 6 x 3 = 18
问题8.通过将4个红色，2个白色和3个绿色标记垂直排列在旗杆上，可以产生多少种不同的信号?
解决方案：
Given: Total number of flags = 9 flags. In which
red appear = 4
white appear = 2
green appear = 3
So, the number of different signals = 9! / (4! 2! 3!) = 1260
问题9.数字1、3、3、0可以形成多少个四位数?
解决方案：
Given: Total number = 4 digits in which
3 appear = twice
Total different numbers = 4! / 2! = 12
But it will also have some 3-digit.
3- digit numbers = 3!/2! = 3
4-digit numbers = 12 – 3 = 9
问题10.“ ARRANGE”一词的字母可以以几种方式排列，以使两个R永远不在一起?
解决方案：
Total permutations of the letters of given word = 7! / (2! 2!) = 1260
Permutations with R together (replace two Rs by 1 letter) = 6! / 2! = 360
Total permutations that satisfies given criteria = 1260 – 360 = 900
问题11：可以用数字0、1、1、5、9形成多少个大于50000的数字。
解决方案：
At first position, only 5 or 9 can be there = 2 ways
Remaining 4 digits can be arranged at 4 positions in 4!/2! ways
Number of different numbers = 2 x 4!/2! = 24
问题12.从“ SERIES”一词的字母开始，可以形成多少个单词，以S开头，以S结尾?
解决方案：
At first and last position, S and S are fixed so 1 way for these two S.
At 2nd, 3rd, 4th, 5th positions = E, R, I, E
so, number of possible arrangements = 4!/2! = 12
So, desired number of such words = 1 x 4!/2! = 12
问题13.“ MADHUBANI”一词的字母中有多少个排列不以M开头但以I结尾?
解决方案：
We fix an I at end so effectively first 8 letters are left to be dealt with
Total number of words that end with I = 8!/2!
Number of words that start with M letter = 7!/2!
Number of desired words = 8!/2! – 7!/2! = 17640
问题14.查找可以用数字2、3、0、3、4、2、3构成的大于一百万的数字数量吗?
解决方案：
To have a number greater than a million it should be at least of 7 length.
So no number starting with digit 0 can be counted.
Total arrangements of given digits are = 7! / (3! 2!) = 420
Arrangements having 0 at first position = 6! / (3! 2!) = 60
Desired arrangements = 420 – 60 = 360