Help me solve this Combination maths problem

This is an IGCSE additional mathematics permutation combination problem which I can't understand. Can somebody explain how this is done? My teacher says the question is incomplete.

Question
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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 sweets are not all of the same flavour, calculate the number of different ways she can select her 3 sweets.

Answer
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3 Different = 6C2 = 20
2 of 1 kind + 1 = 6 * 5 = 30

Total = 2- + 30 = 50

Reply 1

ghostwalker

Original post by Paing Thet

The question is complete, but there is a typo in the answer you've put.

Since the three chosen sweets are not all of the same flavour, either she has chosen 3 different flavours, or she has chosen 2 different flavours.

If 3 different, then she is choosing 3 from 6, and we have 6C3 = 20.

If there are two different flavours, then there is one of one flavour and two of a different flavour.
The one has 6 choices of flavours and once chosen the two have five choices, making 6x5=30 possibilities.

Hence 50 ways in total.

Reply 2

Paing Thet

Original post by ghostwalker

Wow, I understand it now. Thanks

Reply 3

Mathew Pang

When I choose 3 different flavours , why can't it be 6x5x4?

Reply 4

mqb2766

Original post by Mathew Pang

When I choose 3 different flavours , why can't it be 6x5x4?

You'd have over counted as choosing sweets {1,3,6} would be the same as {6,3,1}. You're counting them as different.
For a given arrangement like {1,3,6}, there are 3*2 ways of arranging it, so you'd have to divide your number by 6 (which is 20 different ways as above.).

Reply 5

Mathew Pang

Original post by mqb2766

But how come choose two flavor is 6x5 , but choose three favor is not 6x5x4? If want to ensure it will not repeat, then I guess choose two favor will be (6x5)/2

Reply 6

mqb2766

Original post by Mathew Pang

But how come choose two flavor is 6x5 , but choose three favor is not 6x5x4? If want to ensure it will not repeat, then I guess choose two favor will be (6x5)/2

There are a few ways of explaining it, but assume you choose two different flavours, there are
6*5 / 2
ways of doing this (order not important, divide by 2). There are two possible ways of choosing the 3rd flavour (must be either flavour 1 or flavour 2). Twos cancel, hence
6*5

Reply 7

Robo0master_King

Original post by Paing Thet

This is an IGCSE additional mathematics permutation combination problem which I can't understand. Can somebody explain how this is done? My teacher says the question is incomplete.
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 sweets are not all of the same flavour, calculate the number of different ways she can select her 3 sweets.
Answer
--------------------------------------------------------
3 Different = 6C2 = 20
2 of 1 kind + 1 = 6 * 5 = 30
Total = 2- + 30 = 50

Most of these comments got it right with the 6C3 part, but for the part where you do 2 of 1 kind and 1 of another kind, you could theoretically use permutations. This is because the different ways are 6P2 as you have 112, 223, 334, etc. Technically, you have 6 numbers (1,2,3,4,5,6) and 2 spots as the number is repeated. You have 112 and 211, which is quite similar to 12 and 21. 6P2 is 30, so the answer is 30 + the 20 at the start, which is 50