Kenneth Rosen Edition 7 Exercise 6.3 Question 26 (Page No. 414)

Question

Thirteen people on a softball team show up for a game.

• A. How many ways are there to choose $10$ players to take the field?
• B. How many ways are there to assign the $10$ positions by selecting players from the $13$ people who show up?
• C. Of the $13$ people who show up, three are women. How many ways are there to choose $10$ players to take the field if at least one of these players must be a woman?

Answer 1

$\textbf{(A)}$ We are asked to simply choose any $10$ players out of $13$ to take up the field. Hence, solution to this simply boils down to $^{13}C_{10} = 286$

$\textbf{(B)}$ We can first choose any $10$ players out of $13$ and then arrange them over those $10$ positions, which is effectively given by $^{13}P_{10} = 1,037,836,800$

$\textbf{(C)}$ $13$ people ($3$ women included), we need to form a team of $10$ of which atleast $1$ is a women. One way to solve this is by taking complement, where we subtract the count of no women selected from total number of selections. This would be $^{13}C_{10} - ^{13-3}C_{10} = ^{13}C_{10} - ^{10}C_{10} = 286-1=285$

Answer 2

(a) C(13, 10) = 13!/ 10!3! = 13·12·11/ 1·2·3 = 13 *2 * 11 = 286.

(b) P(13, 10) = 13!/ (13−10)! = 13! /3! = 13 * 12 * 11 * 10* 8* 9 · 8 * 7 * 6 *5 * 4.

(c) If there is exactly one woman chosen, this is possible in C(10, 9)*C(3, 1) = (10! /9!1!)* (3! /1!2!) = 10 * 3 = 30 ways;

two women chosen — in C(10, 8)8C(3, 2) = (10!/ 8!2! )*(3!/ 2!1!) = 45·3 = 135 ways;

three women chosen — in C(10, 7)*C(3, 3) = (10! /7!3!)* (3!/ 3!0!) = 10*9*8*1*2*3 *1 = 120 ways.

Altogether there are 30+135+120 = 285 possible choices.