Question

Q3 X people are chosen from a volley ball team (Take a value of X by yourself, possibly that number must be close to a number of volley ball team members). (2+2+2)
a) How many ways are there to choose Y people to take them to ground(Take value of y by yourself less than x)

b) How many ways are there to assign Z positions by selecting players from X people.(Take Z value by yourself and previous X value.)


c) Of the X people T are women. How many ways are there to choose W players to take them to the field if at least 1 of these players must be a women (take help from example 15)


Hint:
First fill all the values of x,y,z,t and w then your question will be in a mathematical form then you can easily solve them.

Answer 1

Let X = 13, Y = 10, Z = 10, T = 3, W = 3.

(a) $C(13, 10) = \frac{13!}{10!3!} = \frac{13·12·11}{1·2·3} = 13 · 2 · 11 = 286.$

(b) $P(13, 10) = \frac{13!}{(13−10)!} = \frac{13!}{3!} = 13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7 \times 6 \times 5 \times 4.$

(c) If there is exactly one woman chosen, this is possible in $C(10, 9)C(3, 1) = \frac{10!}{9!1!}\frac{3!}{1!2!} = 10 \times 3 = 30$ ways;

two women chosen in $C(10, 8)C(3, 2) =\frac{10!}{8!2!}\frac{3!}{2!1!} = 45 \times3 = 135$ ways;

three women chosen in $C(10, 7)C(3, 3) =\frac{10!}{7!3!}\frac{3!}{3!0!} = 120$ ways.

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