By Kusahara T.

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4. Note that the sample space S consists of 52 5 = 2, 598, 960 possible 5-card hands. 00198079. There are 4 possible suits, and once the suit 52 5 ≥ ¥ is selected, 13 5 possible hands within that suit. 5. 000240096. From 13 kinds, we choose 1. From 52 5 the 4 of this kind, we choose all 4, and from the 48 cards not of this kind, we choose 1. 00144058. There are 13 ways to choose the 52 5 ≥ ¥ kind to get 3 of, and 43 ways to select the three from 4 cards of this kind, and ≥ ¥ There are 12 ways to choose the kind to get 2 of, and 42 ways to select the two ≥ from ¥ ≥ 4¥cards ≥ ¥of≥ this ¥ ≥kind.

Subtracting 2n from the nth term of the original sequence leaves the sequence n2 , so the original sequence is given by an = n2 + 2n . Pn 5. If an = f (n) = i=0 i2 , then the sequence of first diﬀerences is (02 , 12 , 22 , 32 , . ) and thus the sequence of third diﬀerences is constantly 2! = 2. a = 2, we 2 have a = 3! = 13 . Since f (0) = 0 = d, we now have f (n) = 13 n3 + bn2 + cn. From 2 f (1) = 1 = 13 + b + c and f (2) = 12 + 22 = 5 = 83 + 4b + 2c, we find that b = 12 and c = 16 , so f (n) = 12 + · · · + n2 = n3 n2 n n n(n + 1)(2n + 1) + + = (2n2 + 3n + 1) = .

C) The inverse relation {(1,1), (2,1), (3,1), (4,1)} is a function. (d) The inverse relation {(3,1), (4,2), (3,3), (3,4)} is not a function. (e) The inverse relation {(3,1), (1,2), (2,3)} is not a function on {1,2,3,4}. 3. (a) f −1 (x) = (f) f −1 (x) = 3x+1 5 1+2x x−2 4. (b) T = [3, 1), f |−1 T (x) = √ √ x + 4 + 3, or T 0 = (−1, 3], f |−1 T 0 (x) = − x + 4 + 3 (d) T = P({2n − 1|n ∈ N} ∪ {4n|n ∈ N}) and f |−1 T (S) = {s ∈ S|s is odd} ∪ {2s|s ∈ S and s is even}. 0 (e) T = N ∪ {0} and f |−1 T (x) = x − 1, or T = {2n − 1|n ∈ N} ∪ {2 − 2n|n ∈ N} and −1 −1 f |T 0 (x) = x − 1 if x is even and f |T 0 (x) = 1 − x if x is odd.