i could not understand the answer to the question
in how many ways five chocolates can be chosen from an unlimited number of Cadbury, Five Start and Perk chocolates ?
a) 81
b) 243
c) 21
d) 31
answer is b) 243, but according to me should be c) 21.
wud be grateful if anyone can clarify my doubt. also it wud be great if someone can suggest any good resources for P&C and probability(as i am hopelessly weak in these topics).
PS : my first post. :-h this forum has been a great help for me. thank you Forumias. also really appreciate the moderators continued efforts in maintaining the quality of the forum. this is the best place for serious aspirants to share and discuss. Kudos, good job!! :-S
Comments
my approach to this question was that all 5 cud be either cadbury or five star or perk and next either 4 of each and one of the remaining others, and so on. with this approach we get 21 unique combinations.
cadbury five star perk(total = 5)
0 0 5
0 1 4
0 2 3..
....
0 5 0
1 0 4
1 1 3...
...
5 0 0
i wud be grateful if u cud clarify what is wrong with this approach and how the multiple combinations are eliminated(if they have to be eliminated)
2. This problem is not equivalent to filling 5 different slots with the unlimited source of three different entities( say chocolates ) such that each slot get filled by exactly one entity,because the original question is about grouping(only combination)
3. you have arrived at the solution by enumerating every possible instance of the task to be done. An alternative will be to convert it into an another equivalent problem.
4. There can be quite a few equivalent problems like number of ordered non-zero integral pairs that satisfy the equation x+y+z=5 ; or ways of arranging 5 same objects in 3 different slots(say having chocolate brand label) .
5. A very common problem with a cooked up solution is ,ways to arrange objects where not every object is distinct. Say, ways of arranging (a,a,a,a,a,b,b,c,c,c) in linear fashion; solution being 10!/(5!*2!*3!) .Now lets draw an similar equivalent problem for the original question..ways of arranging five "1"s ,and "F" , " S". There is exactly one-to-one relation between answer set to original problem and the equivalent problem. Rather then stating relationship by definition i shall try to convey it through an example:
Take one possible arrangement for the equivalent problem ,say 111F1S1 it wil correspond to (1+1+1=3) chocolate of first label, 1 chocolate of second label, and one chocolate of third labels. I have used 2 markers("F", "S") to separate three imaginary chocalate labels .
Solution to the stated equivalent problem is 7!(5! * 2!) = 21 ,which is indeed the solution to the original problem.
Here you have to partition the number "5" in three parts
i.e. x + y + z = 5
Where x, y, z are any whole number from 0 to 5.
so we can have 0 +0+5 (total 3 ways because 0+5+0 and 5+0+0 are other two ways)
1+4+0 (total 6 ways)
2+2+1 (3 ways)
1+3+1 (3 ways)
2+3+0 (6 ways)
Hence total 21 ways.
x+y+z = k (k=some constant, 5 in above example)
The solution to above problem is (k+n-1) C (n-1) where n is the number of variables (3 in above case, namely x, y, z)...
So the answer to above question becomes -
(5+3-1) C (3-1)
=7C2
=21