Answer:
a) 495
b) 210
c) 35
d) 40
Step-by-step explanation:
Given a total of 12 marbles.
n = 12
Number of red marbles = 7
Number of blue marbles = 5
a) Number of different sets of 4 marbles that can be made from this bag ?
This is a simple combination problem.
where n = 12 and r = 4.
So, answer will be:
[tex]_{12}C_4[/tex]
Formula:
[tex]_{n}C_r = \dfrac{n!}{(n-r)!r!}[/tex]
[tex]_{12}C_4 = \dfrac{12!}{(8)!4!} = \dfrac{12\times 11\times 10\times 9}{4 \times 3\times 2} =\bold{495}[/tex]
b) Two red and two blue marbles:
The answer will be:
[tex]_{7}C_2 \times _{5}C_2 = \dfrac{7\times 6}{2} \times \dfrac{5\times 4}{2} =\bold{210}[/tex]
c) all red marbles.(4 chosen out of 7 red and 0 chosen out of 5 blue marbles)
[tex]_{7}C_4 \times _{5}C_0 = \dfrac{7\times 6\times 5\times 4}{4\times 3\times 2} =\bold{35}[/tex]
d) all red or all blue.(all red marbles plus all blue marbles)
All red marbles:
[tex]_{7}C_4 \times _{5}C_0 = \dfrac{7\times 6\times 5\times 4}{4\times 3\times 2} \times 1=\bold{35}[/tex]
All blue marbles:
[tex]_{7}C_0 \times _{5}C_4 = 1 \times \dfrac{5\times 4\times 3\times 2}{4\times 3\times 2} =\bold{5}[/tex]
So, answer is 40.
