问题15：4本不同的书籍各有三本。它们可以以几种方式排列在架子上?
解决方案：
Total books = 3 x 4 = 12
Ways to arrange books = 12!
Need to compensate for extra ways included due to identically in some books :
The three copies of each book are identical
So, for each different book, they have been included 3! times
So, for the different 4 books, they have been included 3! x 3! x 3! x 3! = 3!4 times
Hence, the ways to arrange the given books on the shelf = 12! / 3!4
问题16.通过使用“数学”一词的所有字母，可以做出多少种不同的安排。其中有多少以C开头?其中有多少以T开头?
解决方案：
Given: In ‘MATHEMATICS’ word
M appear = twice
T appear = twice
A appear = twice
Remaining letters = once
So the total number of letters in the ‘MATHEMATICS’ word = 11
Number of different arrangements = 11! / (2! x 2! x 2!) = 4989600
Beginning with C = fix C on position number 1
Arrange letters at the remaining positions
= 10! / (2! x 2! x 2!) = 453600
Beginning with T = fix T on position number 1
Now, the duplicate of T not left for remaining positions
= 10! / (2! x 2!) = 907200
问题17.研究遗传密码的生物学家有兴趣知道一条链中12个分子可能排列的数目。该链包含以首字母A(对于腺嘌呤)，C(对于胞嘧啶)，G(对于鸟嘌呤)和T(对于胸腺嘧啶)表示的4个不同分子，以及每种分子3个。可能有多少种不同的此类安排?
解决方案：
Given: Total molecules = 12
Ways to arrange molecules = 12!
Need to compensate for extra ways included due to identically in some molecules:
4 kinds of molecules with 3 of each kind = 3! x 3! x 3! x 3! times
Ways to arrange the given books on the shelf = 12!/ 3! x 3! x 3! x 3! = 369600
问题18.如果无法区分相同颜色的光盘，则可以以几种方式连续排列4张红色，3张黄色和2张绿色的光盘?
解决方案：
Given: Total number of discs = 9
In which,
Red color dice = 4
Yellow color dice = 3
Green color dice = 2
So, the total arrangements = 9! / (4! x 3! x 2!) = 1260
问题19.使用数字1、2、0、2、4、2、4可以形成多少个大于1000000的数字?
解决方案：
All 7-digit numbers are greater than 1000000 with digits 1, 2, 0, 2, 4, 2, 4
For 1st digits = number of ways = 6 (except 0)
For 2nd digit = 6 ways (except digit on 1st)
For 3rd digit = 5 ways (except on 1st, 2nd)
And so on.
Number of such numbers = 6 x 6 x 5 x 4 x 3 x 2 x 1 / (3! 2!) (divided to remove duplicates in 2 and 4 digits)
= 6 x 6! / (3! 2!) = 360
问题20.可以用多少种方式排列“暗杀”一词的字母，使所有的S在一起?
解决方案：
Let, all S together can be assumed as 1 symbol = 10 letters left
In the given ASSASSINATION word
A appear = 3 times
N, I appear = 2 times
Number of such ways = 10! / (3! 2! 2!) = 151200
问题21.查找“ INSTITUTE”一词排列的总数。
解决方案：
In the given word INSTITUTE
I appear = 2 times
T appear = 3 times
Hence, the number of permutations are = 9! / (2! x 3!) = 30240
问题22：“ SURITI”一词的字母以所有可能的顺序写成，这些词如在字典中一样被写出。查找单词“ SURITI”的等级。
解决方案：
As we know that in dictionary the words in each stage are arranged in alphabetical order.
According to our problem we can consider that the word begin with I, R, S, T, and U
Number of words starting in I = 5!
Number of words starting in R = 5! / 2!
Number of words starting in SI = 4!
Number of words starting in SR = 4! / 2!
Number of words starting in ST = 4! / 2!
Number of words starting in SUI = 3!
Then the next word in dictionary is going to be = SURIIT
And next = SURITI
So, rank of SURITI = 5! + 5! / 2! + 4! + 4! / 2! + 4! / 2!+ 3! + 1 + 1 = 236
问题23：如果排列了单词“ LATE”的字母并且排列成这样的单词，如在字典中那样，请找到单词LATE的等级。
解决方案：
As we know that in dictionary the words in each stage are arranged in alphabetical order.
According to our problem we can consider that the word begin with A, E, L, and T.
So, Words starting from A = 3!
Words starting from E = 3!
Now, word starting from L = 3!. But one of the word is LATE itself
So the first word is LAET and the next word is LATE
Hence, the rank of LATE is = 3! + 3! + 2 = 14
问题24.如果以所有可能的顺序写上“母亲”一词的字母，并且像在字典中一样写出这些词，请找出“母亲”一词的等级。
解决方案：
As we know that in dictionary the words in each stage are arranged in alphabetical order.
According to our problem we can consider that the word begin with E, H, M, O, T, and R.
So, words starting in E = 5!
Words starting in H = 5!
Words starting in ME = 4!
Words starting in MH = 4!
Words starting in MOE = 3!
Words starting in MOH = 3!
Words starting in MOR = 3!
Words starting in MOTE = 2!
MOTHER = next word
Rank is = 5! + 5! + 4! + 4! + 3! + 3! + 3! + 2! + 1 = 309
问题25.如果将排列a，b，c，d，e一起按字典中的字母顺序记下并编号，则找到排列debac的等级。
解决方案：
As we know that in dictionary the words in each stage are arranged in alphabetical order.
According to our problem we can consider that the word begin with a, b, c, d, and e.
So, the number of words starting in ‘a’ = 4!
Words starting in b = 4!
Words starting in c = 4!
Words starting in da = 3!
Words starting in db = 3!
Words starting in dc = 3!
Words starting in dea = 2!
Next word = debac
Rank = 4! x 3 + 3! x 3 + 2! + 1 = 93
问题26.查找可以将6个“ +”和4个“-”符号排成一行的方式的总数，这样就不会同时出现两个“-”符号。
解决方案：
Total number of ‘-‘ sign = 4
Total number of ‘+’ sign = 6
The six ‘+’ signs arrange in a line = 1 way
Now, we have 7 places in which four different thing can be arranged but all the four ‘-‘ sign look identical
so the ‘-‘ sign can be arranged = 7P4/4! = 35
Hence the number of ways = 1 x 35 = 35
问题27.“中级”一词的字母排列方式可以是多少，以便：
(i)元音总是占据偶数位置吗?
(ii)元音和辅音的相对顺序是否不变?
解决方案：
(i) In the given word, total number of vowels are = 6 vowels
In this word there are total 6 possible even positions
Arrange the vowels on even positions = 6! / (2! x 3!) ways
Arrange consonants = 6! / 2! ways = 360
(ii) Vowels ordering = 6! / (2! x 3!)
Consonants ordering = 6! / 2!
Total permutations = 6! / (2! x 3!) x 6! / 2! = 21600
问题28.“ ZENITH”一词的字母以所有可能的顺序书写。如果所有这些单词都像在词典中一样被写出，则可以有多少个单词? “ ZENITH”一词的等级是多少?
解决方案：
As we know that in dictionary the words in each stage are arranged in alphabetical order.
According to our problem we can consider that the word begin with E, H, I, N, T, and Z.
So, the total words possible are = 6!
Words starting in E = 5!
Words starting in H = 5!
Words starting in I = 5!
Words starting in N = 5!
Words starting in T = 5!
Words starting in ZEH = 3!
Words starting in ZEI = 3!
Words starting in ZENH = 2!
Words starting in ZENIH = 1!
ZENITH is next word
Rank = 5! x 5 + 3! x 2 + 2! + 1 + 1 = 616
问题29.将18只小鼠分为两个实验组和一个对照组，各组的大小均相等。可以通过几种方式将小鼠分为三组?
解决方案：
18 mice can be arranged themselves in 18P18 ways = 18!
According to the question, there are three groups, and they are equally large
So the 18 mice divided into three groups, and they can be arranged themselves inside the group
Hence, the number of ways mice be placed into the three groups = 18!/6! x 6! x 6!
= 18!/(6!)3