Ashish, do note that 12^13^14 is not the same as (12^13)^14. The latter would be 12^182 and thus would behave as you said, but the former would be 12^(13^14) which is 12^(a very large odd number which leaves remainder 1 when divided by 4)…

Oh! Thanks a lot, I get the point. In addition to that, would you please tell me why 12^13^14 is not equal to (12^13)^14 but 12^(13^14). Is it because of some priority of exponents?

You could say it is a property, Ashish. Thing is, when you have an exponent of the form a^b^c, the implicit understanding (default) is that it means a ^(b^c). If you want to do the a^b first, you have to write it specifically as (a^b)^c. Like BODMAS, it is a convention. There has to be a default value, a fallback option.

regards
J

in 4th ques… the formula for napkin ring is 4/3*(pi)h^3… why u r taking h/2

Because in this case, h turns out to be the diameter ðŸ™‚ And we are just finding the volume of a sphere (I am not really looking at it as a napkin ring – one need not know the formula; just the fact that since there is no CBD or NOTA, it MUST be the same answer for ALL possible such figures)

Don’t worry about it too much ðŸ™‚ It is easily the toughest of this entire week’s lot; just included it to show what can be accomplished with a bit of thought and an agile mind.

regards
J

I didnt understand the solution for ques 2 :X
Could you please elaborate that a little more ?
Thanks.

Even though the dice are identical, they need to be treated as different to satisfy the condition of “equiprobability of all cases” so we need to treat them as effectively distinct. This is the same as saying when we toss 3 identical coins the probability of all heads is 1/8 – revise basics of probability to convince yourself…

Hence looking at your cases:
6 6 6 2 => (4!/3!) = 4 ways
6 6 5 3 => (4!/2!) = 12 ways
6 6 4 4 => (4!/2!2!) = 6 ways
6 5 5 4 => (4!/2!) = 12 ways
5 5 5 5 => (4!/4!) = 1 way
The total comes (not surprisingly) as 35.

Still, shouldn’t it be 6*5C3 and not 6*7C3, when we introduce for two variables to be greater than 6, the equation becomes a+b+c+d=6, and we are looking for non-negative solutions?

Sir, In question 2 Is Distributing a shortfall of 4 same as getting a some of 4 which 4 dice are thrown???

But sir , A sum of 4 can be obtained only in 1 way i.e when all dice have 1 appearing on it. But for a shortfall of 4 we have 35 ways.
Isnt both be same??

Oops, I misinterpreted your question! It is not the same as getting a sum of 4, it is the same as distributing 4 among the dice without conditions. The thing is for a “sum of 4” we would have an extra condition of no dice being empty as the minimum possible number is 1. But in the “shortfall of 4” case, a dice can have a shortfall of 0 as well so that condition is relaxed. A shortfall of 4 will be equivalent to getting a sum of 8 I guess.

In the first question,don’t you think 12^13^14 = 12^(4k+2) and will give 12^2 = 144 as reaminder?

Ashish, do note that 12^13^14 is not the same as (12^13)^14. The latter would be 12^182 and thus would behave as you said, but the former would be 12^(13^14) which is 12^(a very large odd number which leaves remainder 1 when divided by 4)…

regards

J

Oh! Thanks a lot, I get the point. In addition to that, would you please tell me why 12^13^14 is not equal to (12^13)^14 but 12^(13^14). Is it because of some priority of exponents?

You could say it is a property, Ashish. Thing is, when you have an exponent of the form a^b^c, the implicit understanding (default) is that it means a ^(b^c). If you want to do the a^b first, you have to write it specifically as (a^b)^c. Like BODMAS, it is a convention. There has to be a default value, a fallback option.

regards

J

in 4th ques… the formula for napkin ring is 4/3*(pi)h^3… why u r taking h/2

Because in this case, h turns out to be the diameter ðŸ™‚ And we are just finding the volume of a sphere (I am not really looking at it as a napkin ring – one need not know the formula; just the fact that since there is no CBD or NOTA, it MUST be the same answer for ALL possible such figures)

regards

J

sorry … i get that.. its modified napkin problem

frankly speaking not able to visualize this problem….

Don’t worry about it too much ðŸ™‚ It is easily the toughest of this entire week’s lot; just included it to show what can be accomplished with a bit of thought and an agile mind.

regards

J

I didnt understand the solution for ques 2 :X

Could you please elaborate that a little more ?

Thanks.

Preeti, I had discussed this earlier when talking of the Theory of Partitioning – see this post and let me know if there is still confusion ( https://crackthecat.wordpress.com/2013/05/20/distribution-with-upper-limits-on-group-size/ ). If need be, visit the 2-3 posts before that for a quick run-through on the Theory of Partitioning.

regards

J

Hello J,

I have doubt about 2nd question.

As all the dice are same.

The cases will be

6 6 6 2

6 6 5 3

6 6 4 4

6 5 5 4

5 5 5 5

I could not think of any more cases.Can you clear?

Thanks,

Even though the dice are identical, they need to be treated as different to satisfy the condition of “equiprobability of all cases” so we need to treat them as effectively distinct. This is the same as saying when we toss 3 identical coins the probability of all heads is 1/8 – revise basics of probability to convince yourself…

Hence looking at your cases:

6 6 6 2 => (4!/3!) = 4 ways

6 6 5 3 => (4!/2!) = 12 ways

6 6 4 4 => (4!/2!2!) = 6 ways

6 5 5 4 => (4!/2!) = 12 ways

5 5 5 5 => (4!/4!) = 1 way

The total comes (not surprisingly) as 35.

regards

J

Dear J,

Just to make myself clear, could u pls tell me in the dice sum why was 6 * 7C3 added again. I am totally lost there.

See this https://crackthecat.wordpress.com/2013/05/20/distribution-with-upper-limits-on-group-size/

We had discussed it in this earlier post (you might also need to read the 3-4 posts prior to that, on Partitioning theory)

regards

J

Still, shouldn’t it be 6*5C3 and not 6*7C3, when we introduce for two variables to be greater than 6, the equation becomes a+b+c+d=6, and we are looking for non-negative solutions?

Sir, In question 2 Is Distributing a shortfall of 4 same as getting a some of 4 which 4 dice are thrown???

Yes it is. Think about it….

regards

J

But sir , A sum of 4 can be obtained only in 1 way i.e when all dice have 1 appearing on it. But for a shortfall of 4 we have 35 ways.

Isnt both be same??

Oops, I misinterpreted your question! It is not the same as getting a sum of 4, it is the same as distributing 4 among the dice without conditions. The thing is for a “sum of 4” we would have an extra condition of no dice being empty as the minimum possible number is 1. But in the “shortfall of 4” case, a dice can have a shortfall of 0 as well so that condition is relaxed. A shortfall of 4 will be equivalent to getting a sum of 8 I guess.

regards

J