So I got this question and we were under the topic of combinations in school. I tried it but first of all here is the question.

A box contains sweets of 6 different flavours. There are AT LEAST 2 sweets of each flavour . A girl selects 3 sweets from the box . Given that these 3 selected sweets are not all of the same flavour , calculate the number of different ways she can select her sweets

my attempt

Since there is 3 sweets needed and they can't be of the same flavour that means 2 must be of one flavour and 1 of the other flavour . So there would be for each flavour, 5 instances where we can have 1 of the selected flavour and 2 of the other flavours. So doing this for each flavour we get, 6(5)=30 and since it can be switched (ie, 2 of selects flavour and 1 of the other flavour) for each flavour we have a total of 30(2) ways in which she can have her sweets. IE 60 ways .

However if this is correct , I want to know how can this be done with combinations ? . We only learnt to select r objects from n distinct objects in class btw ( IE nCr)

  • $\begingroup$ Thnx. Also can you what you did after the 6C3, I am confused about that but I understand where you're saying about me missing each of a single flavour $\endgroup$
    – user122343
    Commented Jul 23, 2018 at 8:38

1 Answer 1


$^{n}C_{r} = \binom{n}{r}$

$\binom{6}{3} + \binom{6}{1} * \binom{5}{1} = 50 $

You forgot about the case of selecting 1 of each flavour.


$\binom{6}{3} : $ Number of ways to choose 3 flavours out of 6 without replacement. Your question only specifies that not all 3 sweets can be of the same flavour. So you also need to consider the case where you pick 3 sweets with all sweets of different flavours.
$\binom{6}{1} * \binom{5}{1} : $ This is the same logic you used expressed in combinatorial form. Out of 6 flavours choose 1 in which to pick 2 sweets from, then choose another to pick the last sweet from. Don't have to multiply by 2 because the reverse case is taken into account already by the first 2 terms.


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