Qualification > Math

Combinatorics

(1/2) > >>

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:

--- Quote from: astarmathsandphysics on October 11, 2009, 12:13:20 pm ---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


--- End quote ---
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:

--- Quote from: astarmathsandphysics on October 11, 2009, 04:04:09 pm ---6*5*4/6 =20 no two the same colour
6*5*2/3=20 two the same colour
40 ways altogether
 

--- End quote ---
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.

Navigation

[0] Message Index

[#] Next page