Collection of questions in combinatorial mathematics

(1) 10 different balls, put in 3 different barrels, how many ways?

Answer: Each ball has 3 kinds of method, ans=310.

(2) Given a string with n characters, suppose all the characters is different from each other, how many different Substrin GS do we have?

Answer: 1 of the length n, n-1 of 2 ... 1 of N, ans=1+2+...+n=n* (n+1)/2. Note: No empty strings are included.

(3) The 6x9 of the square, the beginning of the lower left corner, the end point in the upper right corner, from the beginning to the end, only from the bottom up, from left to right, ask how many different ways to go? Note: The starting/ending points are dots and the side represents the road.

Answer: Need to walk 9 paragraph to the right, walk up 6 paragraphs, each can be up or down, until the direction of the number of segments to go, then ans=c (15,6) = 5005.

(4) Each exam paper has a 8-bit binary serial number. It is only valid if and only if a sequence number contains an even number of 1 o'clock. For example, 00000000, 01010011 are valid serial numbers, and 11111110 are not. So, how many valid serial numbers are there?

Answer: 8 positions, take out an even number of positions to put 1, and all 1 is unordered. Ans=c (8,0) +c (8,2) +c (8,4) +c (8,6) +c (8,8) =1+2*c (8,2) +c (8,4) +1=128.

(5) There are 5 compartments in a row of shelves. There are now 20 books, numbered from 1 to 20. Requires 20 books to be placed in the same row, and numbered from left to right, with at least one book in each lattice, and the numbered 7,8,9 must be in the same lattice. Q, how many possible placement methods are there altogether?

Answer: The 7,8,9 as a book, then the remaining 18 is 1,2...18, a total of 17 gaps, you can put 4 pieces of the board to separate them, into 5 blocks. Ans=c (17,4) = 2380.

(6) The village chief took 4 pairs of father and son to attend the second stop of a village in the third quarter of the shooting. The village in order to protect children not to be abducted have a millennium rule, that is the meal when the child can only be other children or their parents, then 4 pairs of father and son in the Round table total how many kinds of sitting method. (Rotate, everyone faces a change of direction is a new method of sitting)

Answer:-The father, said the child. There are 2 kinds of sitting methods, "- --- "and" - ● - ● - ● - ● ". The first kind, ans1=4!*2*8=384. The second kind: ans2=3*4*8=96. ans=ans1+ans2=480. Note that the second type may be repeated after rotation, the possible situation is: AABBCCDD,AACCDDBB,AACCBBDD (note C father and son next to the substitution), each pair of father and son can also be exchanged with the opposite father and son.

Http://www.nowcoder.com/test/question/done?tid=2012210&qid=15742#summary

Collection of questions in combinatorial mathematics