From Smith, Eggen, & St. Andre, A Transition to Advanced Mathematics, 5th Ed.
Answers to * problems are given in the back of the book and will not be reproduced here.
(pg. 119 : 2, 4, 6, 10, 11, 14, 17, 18, 19 )
2. (a) 8
(c) 13
(d) 33
(e) 23
(f) 36
4.(a) The combination rule.
(b) The product rule.
(c)The combination rule and the product rule.
(d) The permutation rule.
6. (a) C(10,1)C(9,1) = 10(9) = 90.
(b) C(3,1)C(2,1)C(4,1) = 3(2)(4) = 24.
(c) C(3,1)C(4,1)C(5,1) = 3(4)(5) = 60.
10. 7! = 7(6)(5)(4)(3)(2)(1) = 5040.
11. (b) 5(3!)(4!) = 720
(c) 4(4!)(3!) = 576
(d) 4!(3!) = 144 (note that the boys must be in positions 1, 3, 5 and 7).
17. (a) a6 + 6a5b + 15a4b2 + 20a3b3 + 15a2b4 + 6ab5 + b6.
(b) a4 + 8a3b + 24a2b2 + 32ab3 +
16b4.
(c) C(13,3) = 286.
(d) 210C(12,2) = 1024(66) = 676,584.
18.
(b) Consider a set made up of n red objects and m blue objects. If we consider the problem of counting the number of ways to select r of these objects (without regard to color), the answer is C(n+m, r). On the other hand, for any r objects selected, a certain number, say x, of them will be red and the rest, r-x, will be blue. The number of ways to pick x red objects and r-x blue objects is C(n, x)C(m, r-x). If we let x vary from 0 to r, we will again be counting all the possible ways of selecting r objects, so, C(n+m, r) = Sum {x = 0 to r} C(n,x)C(m, r-x).
(c) From the pascal triangle relation we have, C(2n, n) + C(2n, n+1) = C(2n+1, n+1). Consider the formula for C(2n+2, n+1) = (2n+2)!/(n+1)!(n+1)! = (2n+2)(2n+1)!/(n+1)!(n+1)(n!) = [(2n+2)/(n+1)][(2n+1)!/(n+1)!n!] = 2C(2n+1, n+1). So, C(2n+1, n+1) = ½C(2n+2, n+1). We can give a combinatorial proof of the second part of this. Consider a set with 2n+2 elements and distinguish one of the elements of this set (think of it as being colored red). The number of subsets of size n+1 is C(2n+2, n+1). This must be an even number because these subsets come in pairs since the complement of a subset of size n+1 is also of size n+1. The red element is in exactly one of each of these pairs, so the number of n+1 subsets which do not contain the red element is ½C(2n+2, n+1). On the other hand, we would be counting the same collection of subsets if we removed the red element (leaving 2n+1 elements) and then counted all the subsets of size n+1, i.e., C(2n+1, n+1). Thus, C(2n+1, n+1) = ½C(2n+2, n+1).
19. (a) A.
(b) A.
(c) A.