328 ALGEBRA. 400. To find the total number of ways in which it is possible to make a selection by taking some or all of n things.
Each thing may be dealt with in two ways, for it may either be taken or left; and since either way of dealing with any one thing may be associated with either way of dealing with each one of the others, the number of ways of dealing with the n things is
2 x 2 x 2 x 2 .. to n factors.
But this includes the case in which all the things are left, therefore, rejecting this case, the total number of ways is
2^n – 1.
This is often spoken of as "the total number of combinations" of n things.
Ex. A man has 6 friends; in how many ways may he invite one or more of them to dinner?
He has to select some or all of his 6 friends; and therefore the number of ways is 2^6 – 1, or 63.
This result can be verified in the following manner.
The guests may be invited singly, in twos, threes, number of selections; therefore the
=6C1 + 6C2 + 6C3 + 6C4 + 6C5 + 6C6 = 6 + 15 + 20 + 15 + 6 + 1 = 63.
401. To find for what value of r the number of combinations of n things r at a time is greatest.
Since nCr = n (n – 1) (n – 2) ... (n – r + 2) (n – r + 1) 1.2.3... (r – 1)r
and nCr-1 = n (n – 1) (n – 2) ... (n -- r + 2) 1.2.3... (r –1)
nCr = nCr-1 x n–r+1 r
The multiplying factor n-r+1 r may be written n+1 r -1, which shows that it decreases as r increases. Hence as r receives the values 1, 2, 3, ... in succession, nCr is continu-
