ix.] COMBINATIONS AND PERMUTATIONS. 187 



line. The combinations of three things out of seven are 

 -* or 35, which appears fourth in the eighth line. 



1X2X3 



In a similar manner, in the fifth, sixth, seventh, and eighth 

 columns of the eighth line I find it stated in how many 

 ways I can select combinations of 4, 5 , 6, and 7 things out 

 of 7. Proceeding to the ninth line, I find in succession 

 the number of ways in which I can select I, 2, 3, 4, 5, 6, 

 7, and 8 things, out of 8 things. In general language, if 

 I wish to know in how many ways m things can be 

 selected in combinations out of n things, I must look in 

 the n + I th line, and take the m + I th number, as the 

 answer. In how many ways, for instance, can a sub- 

 committee of five be chosen out of a committee of nine. 

 The answer is 126, and is the sixth number in the tenth 



line: it will be found equal to 9 " '-^ *. which 



1.2.3.4.5' 



our formula (p. 182) gives. 



The full utility of the figurate numbers will be more 

 apparent when we reach the subject of probabilities, but I 

 may give an illustration or two in this place. In how 

 many ways can we arrange four pennies as regards head 

 and tail ? The question amounts to asking in how many 

 ways we can select o, i, 2, 3, or 4 heads, out of 4 heads, 

 and the fifth line of the triangle gives us the complete 

 answer, thus 



We can select No head and 4 tails in I way. 

 I head and 3 tails in 4 ways. 



2 heads and 2 tails in 6 ways. 



3 heads and I tail in 4 ways. 



4 heads and o tail in I way. 



The total number of different cases is 16, or 2 4 , and 

 when we come to the next chapter, it will be found that 

 these numbers give us the respective probabilities of all 

 throws with four pennies. 



I gave in p. 1 8 1 a calculation of the number of ways in 

 which eight planets can meet in conjunction ; the reader 

 will find all the numbers detailed in the ninth line of the 

 arithmetical triangle. The sum of the whole line is 2 8 or 

 256; but we must subtract a unit for the case where no 

 planet appears, and 8 for the 8 cases in which only one 

 planet appears ; so that the total number of conjunctions 



