College algebra by A. Adrian Albert

By A. Adrian Albert

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IV. using the numbers 3, 4, 5, 6, 7, 9 as digits? Remarks: The problem involves arrangement and is a permutation problem. We shall give two solutions one of which uses the exclusion principle. Solution 1 With the six digits it is possible to form PM four-digit numbers. 8 end in 4 and the same amount in 6. They are to be omitted. 8 = (6' 5 . 4 . 3) - (2' 5 . 4 . 3) = (6 - 2)(5' 4 . 3) = 240. Solution 2 The final digit is permitted to be anyone of four numbers. Mter it has been selected, the remaining three digits are to be selected and arranged from the remain,ing five numbers.

In the particular case where s = n - r, the value of formula (29) is C""r and we have shown that the number of distinguishable permutations of n = r + s objects of which r are alike and the remaining s alike is C""r = C",,8' Suppose finally that we write n as a sum (30) n of positive integers. of n objects into = nl + ... i objects are alike, formula (31) gives a formula for the number of distinguishable permutations. n21 . • • n, I. Let us observe, in closing, that there is no formula for the number of distinguishable permutations of n objects r at a time if some of the objects are alike.

2. Am. 33. (k) 1648, 2997} (l) {894,3278} Am. 298. Am. 12. (m) 11442, 6489} (~) Am. 49. Am. 16. (0) (p) (q) (r) {l432, 8469} 11134, 8019} 11608,17523} 1768, 9552} 1266664, 877769} Am. 1. Am. 3. Am. 11,111. 42 [CHAP. INTEGER,S 'II 2. Use the method of Inustrative Example II to show that (a) {21924,2144) = 4 (d) {5103, 7614) = 81 (e) {4526,9344) = 146 (f) {7992, 17640} = 72 (b) {179,21) = 1 (c) {4078, 814) = 2 8. Linear combinations ar'e integers, the sum (FULL COUR,SE). If a, b, m, n, ma+nb is called a linear combination of a and b.

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