Combinatorics

[astarmathsandphysics]

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 sweet are not all of the same flavouor, calculate the no. of different ways she can select her 3 sweets.

6C3 ways of secting 3 sweets altogether but 6 of these have all same colour so ans=6C3-6

[slvri]

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 sweet are not all of the same flavouor, calculate the no. of different ways she can select her 3 sweets.

6C3 ways of secting 3 sweets altogether but 6 of these have all same colour so ans=6C3-6

the answer should be 50

[astarmathsandphysics]

6*5*4/6 =20 no two the same colour
6*5*2/3=20 two the same colour
40 ways altogether
 

[slvri]

6*5*4/6 =20 no two the same colour
6*5*2/3=20 two the same colour
40 ways altogether
 

6C3=20 no two of the same colour
6*5=30 two of the same colour
total=30+20=50
when i gave my teacher this q he couldnt solve it

[astarmathsandphysics]

my reasong for 6*5*2/3 is this: For the first you have a choice of 6 and for the second a choice of 5. For the tird you can have the first or second colour, hence the factor of 2 and divide by three because the odd colour can be 1st, 2nd, or third.

[slvri]

my reasong for 6*5*2/3 is this: For the first you have a choice of 6 and for the second a choice of 5. For the tird you can have the first or second colour, hence the factor of 2 and divide by three because the odd colour can be 1st, 2nd, or third.

but couldnt the logic also be this: for the first and second sweets you have a choice of 6 and for the third sweets you have a choice of 5 so 6*5=30
im confused......

[astarmathsandphysics]

The sweets can be picked in any order and the order doesnt matter. Things get much worse because you can have two sweets the same colour but not three.