By Joe L. Mott
ISBN-10: 0835913910
ISBN-13: 9780835913911
Ebook by way of Mott, Joe L., Kandel, Abraham, Baker, Theodore P.
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This result of course coincides with that obtained in the first solution. It shows that among the numbers under consideration there are more with I's among their digits than without. = - 52 SOLUTIONS llb. Among the integers from I to 222,222,222 there are 22,222,222 ending in a 0 (namely, the numbers 10,20,30, ... , 222,222,220). In order to determine how many integers have a 0 in the next to last position, notice that what comes before this 0 can be anything from I to 2,222,222, while what comes after it can be anything from 0 to 9.
Then g is counted by each term of (I), and is therefore counted a net of I + I + I - I - I - I + I = I time. This analysis shows that expression (I) counts each element of A V B V C once. On the other hand, elements not in A V B V C are not counted in any of its terms, and therefore (I) is equal to #(A V B V C). 12c. The general case can be treated by the same reasoning as that used in part b. We must show that in the expression #(A 1) + #(AJ + ... + #(Am) - #(Al (\ AJ - #(Al (\ Aa) - ... - #(A mAm) + #(Al (\ A2 (\ Aa) + ...
Suppose that as N -- 00 the ratio q(N)IN approaches a limit; in this case this limit is called the probability that a number selected at random from the entire sequence has the desired property. Note that this probability depends on the way in which the numbers are arranged in a sequence. Changing the order of the numbers can change the value of the probability. Example: consider the positive integers arranged in increasing order: 1,2,3, ... Of the first N of these numbers, [N/2] are even; as N -- 00 the ratio [NI2]/N approaches i, which means that the probability that any number selected at random is even equals 1/2.
Discrete Mathematics For Computer Scientists And Mathematicians by Joe L. Mott
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