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Unformatted text preview: ECE 190 — Fall, 2006 Assignment 7 Solutions Section 6.3 • 5(a) To be divisible by five, the last digit must be 0 or 5. The first digit cannot be zero so there are 9 × 10 × 10 × 10 × 2 = 18 000 such integers. • 11 See text. • 21 (a) See text. • 21 (c) From part (a), there are 1000 357 = 643 such integers. • 28 See text. Section 6.5 • 3 See text. • 5 See text. • 10 See text. • 11 See text. • 14 Start by taking 50 of the units and distributing them so that each of a, b, c, d, e has 10 units. The number of ways of distributing the remaining 450 units is equal to the number of ways of arranging 450 × ’s and 4  ’s. This is 454! 450!4! = 1 746 858 751 • 16(a) We can divide 30 coins into 4 groups (pennies, nickels, dimes, and quarters) using an arrange ment of 30 × ’s and 3  ’s. The number of possibilities is 33! 30!3! = 5456 • 17 Since the sum of the digits is to be 9, we can represent the sum by 9 × ’s. We need 5 digits to add to 9 so we can break up the digits into 5 groups (1st digit, 2nd digit, ... , 5th digit) usingadd to 9 so we can break up the digits into 5 groups (1st digit, 2nd digit, ....
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This homework help was uploaded on 04/19/2008 for the course ECE 190 taught by Professor Carter during the Fall '06 term at University of Toronto.
 Fall '06
 Carter

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