5 5 Sisters 4 4 Cookies

36 36 30 30 34 34 32 32

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5 solutions

Sayonto Khan
Mar 20, 2014

Easy Solution...!

first thinking of 1 girl missing the treat, then how many possibilities will be there ? = 5C4 then, thinking of 2 girls missing the treat, then how many possibilities will be there ? = 5C3 then, thinking of 3 girls missing the treat, then how many possibilities will be there ? = 5C2 then, thinking of 4 girls missing the treat, then how many possibilities will be there ? = 5C1

each of the girls cannot miss the cookies, otherwise the plate wouldn't have been empty, would it ?

so, add every possible solution, you get the result....!

represent whether a girl had a cookie or not by either a zero or a 1. Then the set of all possible representations = 2^5 = 32. Of this all zero = 00000 and all ones 11111 are not possible. All the rest are possible, and hence the total number of possibilities = 32-2 = 30.

Nidhin Koshy - 7 years, 2 months ago

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I thought it out Sayonto's way.....but yours is pure genius...well done!!!

Tanya Gupta - 7 years, 2 months ago

Indirectly u did the same thing as Sayonto did. Bcoz 5C1 + 5C2 + 5C3 + 5C4 = (5C0 + 5C1 + 5C2 + 5C3 + 5C4 + 5C5) - (5C0 + 5C1) = 2^5 - (5C0 + 5C1) = 2^5 - 2 = 30 Anyway sir, urs process is a nice short way to solve it.

Amlan Mishra - 7 years, 2 months ago

Brilliant

Lee Nguyen - 7 years, 2 months ago

the problem doesn't specify if the cookies are distinguishable.

Frodo Baggins - 7 years, 2 months ago

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Ya, thats why the there are 30 such ways.

Amlan Mishra - 7 years, 2 months ago

Very good and clear solution, But I am having problems figuring out what went wrong with mine. Would be of great help if anyone cared to explain. 4 cookies can be distributed in the following ways: 4 alone --- 5C1 3, 1 --- 5C2 2, 2 --- 5C2 2, 1, 1 --- 5C3 1, 1, 1, 1 --- 5C4 Total combos = 5C1 + 5C2 + 5C2 + 5C3 + 5C4 = 40

Aushik Barua - 7 years, 2 months ago

I disagree. Every girl is a different person so isn't it suppose to be a permutation? After all, the question did mention "with respect to who ate and who did not".

Shaun Loong - 7 years, 2 months ago

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As in the question, it is mentioned "with respect to who ate and who did not", There are 5C1 ways in which 1girl does not eat; 5C2 ways in which 2 of the 5 girls do not eat; 5C3 ways in which 3 of the 5 girls do not eat; 5C4 ways in which 4 of the 5 girls do not eat. So the required solution is 5C1 + 5C2 + 5C3 + 5C4 = 5 + 10 + 10 + 5 = 30

Amlan Mishra - 7 years, 2 months ago

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Ahhhhh..I see it now! Thanks! :D

Shaun Loong - 7 years, 2 months ago

Each girl either ate one or more cookies, or did not. This gives 2^5 = 32 possibilities. There are two possibilities that must be discounted: all five could not eat cookies (there are only four) and all five could not have eaten cookies (all four cookies were eaten). Thus, the possibilities are 32 - 2 = 30.

Solayman Khan
Mar 31, 2014

5C1+5C2+5C3+5C4=30

Anand Shah
Mar 26, 2014

easy to solve... count 5C4 if it will be classified among 4 count 5C3 if it will be classified among 3 count 5C2 if it will be classified among 2 count 5C1 if it will be classified among 1 total of 5C1+5C2+5C3+5C4+5C5=5+10+10+5=30

Amlan Mishra
Mar 25, 2014

5C1 + 5C2 + 5C3 + 5C4 = 5 + 10 + 10 + 5 = 30

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