How many toys can be chosen?

Question

There are 3 red, 5 blue, 2 yellow and 4 green toys in the box. In how many different ways can 6 toys be chosen if one of them should be blue and the other one - yellow? I came up with a solution but i am not sure if it is right. 2 toys out of 6 should be of a specific colour. Then, only 4 toys can be of any other colour. so, the number of different ways $= C^4_{12} = \frac{12!}{4!8!}=495$

Answer 1

Given the clarifications in comments – the toys are distinct, order doesn't matter, exactly one is blue and exactly one is yellow – we have

• $5$ ways to choose the blue toy and $2$ ways to choose the yellow one
• $\binom74=35$ ways to choose the other four toys

Thus there are actually $5×2×35=350$ valid selections.

Answer 2

So, we want to count how many ways we can select six toys where at least one was blue and at least one was yellow where all toys are distinct and order of selection doesn't matter. (For example, getting three blue toys, two yellow toys and one red toy is acceptable)

To do this, let us approach indirectly by instead counting how many ways we can do this if we don't care about the restriction of getting at least one blue and at least one yellow, and subtract from this the number of selections where we violate either condition.

The number of ways to do this without restriction is $\binom{14}{6}$.

The number of ways to violate at least one condition we use inclusion-exclusion. First counting the number of ways we violate the first condition, then counting the number of ways we violate the second condition, and finally correcting our count by removing those ways we overcounted where we violate both conditions.

Getting no blues can be done in $\binom{9}{6}$ ways. Getting no yellows can be done in $\binom{12}{6}$ ways. Getting neither blue nor yellow can be done in $\binom{7}{6}$ ways. This gives a total of $\binom{9}{6}+\binom{12}{6}-\binom{7}{6}$ ways in which we violate at least one condition of selecting at least one blue and at least one yellow.

Our final count then:

$$\binom{14}{6}-\binom{9}{6}-\binom{12}{6}+\binom{7}{6}$$

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Your attempt was incorrect for two reasons. First, it neglected to take into account which blue was actually chosen in the first step and which yellow was chosen. Your calculations would not have been able to distinguish between certain arrangements. To see this error in action, consider the problem where we have only $2$ blues and $1$ yellow toy where we select two toys, at least one needing to be blue and at least one needing to be yellow. It should be obvious there are two possible outcomes, but your attempt would have you believe there is only one possibility.

To correct this, multiply by the number of blues at the start and the number of yellows. This would have put your total to $5\times 2\times \binom{14}{4}$. But, correcting it like this incorrectly puts significance on which blue was the "guaranteed" blue selected in the first step in the case that there is at least one blue chosen in the later step. To see this error in action... consider the problem where we only have blue toys. If we have $2$ blue toys and we want to pick two of them while making sure we have at least one blue. The answer is obviously going to be $1$, no calculations required, but if we were to follow your logic (after modifications) there would be $2\times \binom{1}{1}=2$

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For the problem where we want exactly one blue and exactly one yellow toy, (For example, getting three blue toys, two yellow toys and one red toy is not acceptable), see @ParclyTaxel's answer.