ON DIRECT PROBABILITIES. 35 



Thus, if I ask how many combinations of 21 can be taken 

 out of 25, I do in effect ask how many combinations 

 of 4 may be taken. For there are just as many ways 

 of taking 2 1 as there are of leaving 4. 



The following table corresponds to the one preced- 

 ing : 



Thus the combinations of 5 out of 9 are 126 in 

 number. 



3. When the same event may be repeated in a per- 

 mutation,, the number of permutations is the product 

 of as many numbers, all equal to the whole number of 

 possible events, as there are of events in each permutation. 

 Thus, of four events, A, B, C, D, the number of per- 

 mutations of two, with repetition, is 4 x 4 or 1 6 ; with- 

 out repetition 4 x 3 or 12. The additional 4 in the 

 first case, are A A, B B, C C, and D D. The number 

 of permutations of three, with repetition, is 4 x 4 x 4, 

 or 64 without repetition, 4x3x2 or 24. Of the addi- 

 tional 40, 10 are where A only is repeated; namely, A A B, 

 A A C, A AD, A B A, A C A, A D A, B A A, C A A, 

 D 2 



