0
$\begingroup$

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)

$\endgroup$
1
  • $\begingroup$ 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

0
$\begingroup$

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

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

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

EDIT:

$\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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.