Question

A box contains 5 red and 4 blue balls. in how many ways can 4 balls be chosen such that there are at most 3 balls of each colour?

Answer 1

Hello!

A box contains 5 red and 4 blue balls. in how many ways can 4 balls be chosen such that there are at most 3 balls of each colour?

We have a simple combination, whose formula is as follows:

$C_{n,r} = \dfrac{n!}{r!(n-r)!}$

If: there are at most 3 balls of each colour...

[First Step] Let's apply the formula, (to 3 red balls and 1 blue ball), let's see:

$C_{n,r} * C_{n,r} =\:?$

$C_{5,3} * C_{4,1} =\:?$

$\dfrac{5!}{3!(5-3)!} * \dfrac{4!}{1!(4-1)!} =\:?$

$\dfrac{5!}{3!2!} * \dfrac{4!}{1!3!} =\:?$

$\dfrac{5*4*3*\diagup\!\!\!\!2!}{3!\diagup\!\!\!\!2!} * \dfrac{4*\diagup\!\!\!\!3!}{1*\diagup\!\!\!\!3!} =\:?$

$\dfrac{5*4*3}{3*2} * \dfrac{4}{1} =\:?$

$\dfrac{60}{6} *4 =\:?$

$10 * 4 = \boxed{40\:ways}$

[Second Step] Let's apply the formula, (to 2 red balls and 2 blue balls), let's see:

$C_{n,r} * C_{n,r} =\:?$

$C_{5,2} * C_{4,2} =\:?$

$\dfrac{5!}{2!(5-2)!} * \dfrac{4!}{2!(4-2)!} =\:?$

$\dfrac{5!}{2!3!} * \dfrac{4!}{2!2!} =\:?$

$\dfrac{5*4*\diagup\!\!\!\!3!}{2!\diagup\!\!\!\!3!} * \dfrac{4*3*\diagup\!\!\!\!2!}{2!\diagup\!\!\!\!2!} =\:?$

$\dfrac{5*4}{2} * \dfrac{4*3}{2} =\:?$

$\dfrac{20}{2} * \dfrac{12}{2} =\:?$

$10 * 6 = \boxed{60\:ways}$

[Third Step] Let's apply the formula, (to 1 red ball and 3 blue balls), let's see:

$C_{n,r} * C_{n,r} =\:?$

$C_{5,1} * C_{4,3} =\:?$

$\dfrac{5!}{1!(5-1)!} * \dfrac{4!}{3!(4-3)!} =\:?$

$\dfrac{5!}{1!4!} * \dfrac{4!}{3!1!} =\:?$

$\dfrac{5*\diagup\!\!\!\!4!}{1!\diagup\!\!\!\!4!} * \dfrac{4*\diagup\!\!\!\!3!}{\diagup\!\!\!\!3!} =\:?$

$\dfrac{5}{1} * 4 =\:?$

$5 * 4 = \boxed{20\:ways}$

Therefore, to 3 balls of each colour, we have:

$C_{5,3} * C_{4,1} + C_{5,2} * C_{4,2} + C_{5,1} * C_{4,3} =\:?$

$40 + 60 + 20 =\:?$

$Answer = \boxed{\boxed{\boxed{120\:ways}}}\end{array}}\qquad\checkmark$

Answer:  

120 ways

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I Hope this helps, greetings ... Dexteright02! =)