Tham khảo tài liệu 'a course in mathematical statistics phần 2', ngoại ngữ, ngữ pháp tiếng anh phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 38 2 Some Probabilistic Concepts and Results Then by Theorem 6 there are 13-6 220 -64 1 098 240 poker hands with one pair. Hence by assuming the uniform probability measure the required probability is equal to 1 098 240 . . 2 598 960 . THEOREM 9 i The number of ways in which n distinct balls can be distributed into k distinct cells is kn. ii The number of ways that n distinct balls can be distributed into k distinct cells so that the jth cell contains n balls n 0 j 1 . k n n is n _ n 1 n2 nk 1 1 n2 . nk J. iii The number of ways that n indistinguishable balls can be distributed into k distinct cells is kA k n -1 I n J- Furthermore if n k and no cell is to be empty this number becomes n -1 . I k - 1J PROOF i Obvious since there are k places to put each of the n balls. ii This problem is equivalent to partitioning the n balls into k groups where the jth group contains exactly n balls with n as above. This can be done in the following number of ways nTn 44n- n l I n JI n2 JI nk J n n1 n2 nk iii We represent the k cells by the k spaces between k 1 vertical bars and the n balls by n stars. By fixing the two extreme bars we are left with k n -1 bars and stars which we may consider as k n - 1 spaces to be filled in by a bar or a star. Then the problem is that of selecting n spaces for the n stars which can be done in r -1 ways. As for the second part we now have the condition that there should not be two adjacent bars. The n stars create n - 1 spaces and by selecting k - 1 of them in J ways to place the k - 1 bars the result follows. Combinatorial Results 39 REMARK 5 i The numbers n j 1 . k in the second part of the theorem are called occupancy numbers. ii The answer to ii is also the answer to the following different question Consider n numbered balls such that n are identical among themselves and distinct from all others nj 0 j 1 . k ĩ k 1 nj n. Then the number of different permutations is 1 n 1 . ni n2 . nk J Now consider the following examples for the purpose
