54 thoughts on “Distribution with upper limits on Group Size”

sir, in the last example alternate solution works. same alternate solution works for second last question (x+y+z=12. none exceeding 5).
but, why is it not working in case of a+b+c+d+e+f=13. none exceeding 9.
i got the traditional method, my query is what’s the restriction for alternate method discussed by you.

The balances should not exceed the total in the alternative method. For example when we were looking at a +b +c = 130, we were looking at x + y + z = 20, where x = 50 – a, y = 50 – b and z = 50 – c so there was no way any of those three could cross 50 (thus making a or b or c negative).

However with a+b+c+d+e+f=13, none exceeding 9, we get the corresponding equation as u + v + w + x + y + z = 9*6 – 13 = 41, and again u, v, w etc can be more than 9 so we would still have a cumbersome solution. As long as the RHS of the equation exceeds 9, we have to do extra work.

But if we had, for example, a+b+c+d+e+f = 47, none exceeding 9, then we could solve it as u + v + w + x + y + z = 9*6 – 47 = 7 which gives us 12C5 ways.

“The balances should not exceed the total in the alternative method..” a+b+c+d+e+f=13.. none exceeding 9.. what is “the total” here 9 or 13. as u mentioned in second para, i think its 9.. isn’t it

can we conclude that when balance is maintained and here, like none can exceed 9, we are sure, so in these kind of cases, double counting will never occur, we would never be subtracting any cases (as did in a Q5)?

I have no idea what that question meant, I’m afraid! But if the total is less than 18, then of course two digits cannot cross 9 together (if that is what you’re asking).

regards
J

yes ..thanks..

Anand, here the balance requirement is 54 – 13 = 41. And that exceeds 9.

It is like a Venn diagram…when we counted the cases of A exceeding the limit and those of B exceeding the limit, the cases of both A and B exceeding the limit were counted twice. So we have subtracted them twice and need to add them back…

Dear Sir i am not getting why are you adding again please, explain by a venn diagram picture. Upload it somewhere and share the picture. Please i ve tried but not getting a clear picture!!

Divya, consider the case (x, y, z) = (6, 6, 0). This would be subtracted when we apply the condition x > 5 as well as when we apply the condition y > 5, right? So that is an extra case being subtracted as the same case is being removed twice.

Hi.. is it possible to apply the logic used in the first question to one with eight digits whose sum of digits is 4? I have tried but I seem to be messing it up somewhere

Yes you can. Basically remember that the first digit cannot be 0. So allot one there. Now the remaining 3 can go to any of the 8 places in 3+8-1C8-1 = 10C7 = 120 ways

In this case, since 5 is missing, it is not “all numbers” so it would probably actually be simpler to do it manually. If you wanted to do it using PnC you would need to figure how to distribute 6 runs (shortage from the max of 36) among 6 balls such that they could be 0, 2, 3, 4, 5, or 6…which is rather a pain!

Hi.. in the question x+y+z<=12 with x,y,z<=5 what would be the method if there is an upper limit only for 2 variables say x&y <=5 and z have no restriction?

Hi.. in the question x+y+z<=12 with x,y,z<=5, what would be the method if there is an upper limit only for 2 variables say x,y<=5 and z has no restrictions?

Again, there would be cases where you are taking away more than 7 fish from a single person in that 27C4. In fact you will end up having more work, as when removing 23 fishes, there could be as many as 3 people who exceed the limit!

First of all why would you do this? If I take the complement it is only to make life more simple. If the new number 23 is bigger than the original 12, then no point taking it.

Secondly, a’, b’ etc are not the fish, they are the shortage of fish. So if I tell you that in 3 subjects, out of 100 each you scored 110 marks. Then the marks you missed out on would be 300 – 110 = 190 right? i.e. if a + b + c = 110 then a’ + b’ + c’ = 190. That is what we are writing in the given problem.

Yea. Thanks J :). Just another question J. If there are six cities which are to be connected by wires with the constraint that each city can have exactly incoming 2 wires from different cities in how many ways can one connect?

I am assuming that it is fine if we form two separate and disconnected closed loops. So then we have two options. (If not, only the first one)

Single ABCDEF loop which can be done in 5!/2 ways or Two loops ABC and DEF which can be done in 6C3 ways. (Do recheck the numbers as I did them in haste and may have made some mistake!)

sir reg. 5!/2, the concept i used was that, a city can choose 2 out of 5 and so 6 cities doing the same will give me 5!*6 which gives me the same ans. but am i correct in doing so?

Hello Sir..in the first example..a + b = 13 , but a cannot be 0, so a’ + b = 12, which gives 13 cases without any constraints..this is not considered in the example. I know something is wrong here as through manual calculation, the answer is correct (6). Can you please tell me what I am doing wrong

Please correct me if I am wrong..is the reason we are not considering a not equal to 0 in the initial case, because it will be neglected by taking out the values of b greater than 10?

Hello sir,
In Question 5, the fishing problem, according to the method new equation is a’ + b + c + d + e = 5 and there are 5 farmers so shouldn’t the answer should have 9C4 instead of 8C4 ?

Sir, why arent we including the case where any two person can carry away with more than 7 fishes just like we are doing in the example post it, where any 2 person can sum upto more than 7? Does that means that will still effectively leave us with 5 to be distributed among rest 3, thus they;d be obvio below 7, that’s why we excluded those cases?

sir, in the last example alternate solution works. same alternate solution works for second last question (x+y+z=12. none exceeding 5).

but, why is it not working in case of a+b+c+d+e+f=13. none exceeding 9.

i got the traditional method, my query is what’s the restriction for alternate method discussed by you.

The balances should not exceed the total in the alternative method. For example when we were looking at a +b +c = 130, we were looking at x + y + z = 20, where x = 50 – a, y = 50 – b and z = 50 – c so there was no way any of those three could cross 50 (thus making a or b or c negative).

However with a+b+c+d+e+f=13, none exceeding 9, we get the corresponding equation as u + v + w + x + y + z = 9*6 – 13 = 41, and again u, v, w etc can be more than 9 so we would still have a cumbersome solution. As long as the RHS of the equation exceeds 9, we have to do extra work.

But if we had, for example, a+b+c+d+e+f = 47, none exceeding 9, then we could solve it as u + v + w + x + y + z = 9*6 – 47 = 7 which gives us 12C5 ways.

regards

J

“The balances should not exceed the total in the alternative method..” a+b+c+d+e+f=13.. none exceeding 9.. what is “the total” here 9 or 13. as u mentioned in second para, i think its 9.. isn’t it

can we conclude that when balance is maintained and here, like none can exceed 9, we are sure, so in these kind of cases, double counting will never occur, we would never be subtracting any cases (as did in a Q5)?

I have no idea what that question meant, I’m afraid! But if the total is less than 18, then of course two digits cannot cross 9 together (if that is what you’re asking).

regards

J

yes ..thanks..

Anand, here the balance requirement is 54 – 13 = 41. And that exceeds 9.

regards

J

thanks sir…

I dont get why we are adding again for tow out of three variable is 0. In last sum 3*30c2. Plz help me out sir..!

It is like a Venn diagram…when we counted the cases of A exceeding the limit and those of B exceeding the limit, the cases of both A and B exceeding the limit were counted twice. So we have subtracted them twice and need to add them back…

regards

J

In the fishing problem mentioned above when subtracting the cases i got 9c4.Please explain elaborately.

Dear Sir i am not getting why are you adding again please, explain by a venn diagram picture. Upload it somewhere and share the picture. Please i ve tried but not getting a clear picture!!

When the case x greater than 5 and y greater than 5 and z greater than 5 how can it be counted twice. Please shed some light in that.

Divya, consider the case (x, y, z) = (6, 6, 0). This would be subtracted when we apply the condition x > 5 as well as when we apply the condition y > 5, right? So that is an extra case being subtracted as the same case is being removed twice.

regards

J

Hi.. is it possible to apply the logic used in the first question to one with eight digits whose sum of digits is 4? I have tried but I seem to be messing it up somewhere

Yes you can. Basically remember that the first digit cannot be 0. So allot one there. Now the remaining 3 can go to any of the 8 places in 3+8-1C8-1 = 10C7 = 120 ways

regards

J

In Cricket, a batsman can score 0, 1, 2, 3, 4 or 6 runs of a ball. What is the number of distinct

sequences in which exactly 30 runs can be scored in an over of six balls? Assume that all the runs

are scored by batsmen only and there are no extra balls/runs.

In this case, since 5 is missing, it is not “all numbers” so it would probably actually be simpler to do it manually. If you wanted to do it using PnC you would need to figure how to distribute 6 runs (shortage from the max of 36) among 6 balls such that they could be 0, 2, 3, 4, 5, or 6…which is rather a pain!

regards

J

Hi.. in the question x+y+z<=12 with x,y,z<=5 what would be the method if there is an upper limit only for 2 variables say x&y <=5 and z have no restriction?

Hi.. in the question x+y+z<=12 with x,y,z<=5, what would be the method if there is an upper limit only for 2 variables say x,y<=5 and z has no restrictions?

Then we will get 2 x 8C2 and 1 x 2C2 instead. Think about it….

regards

J

How many non negative solutions exist for x=+y+z=12 such that none of x,y,z exist 3

exceeds 3 not exist typo

Would be 0? If none exceeds 3, the sum can’t exceed 9…

regards

J

what is this =+ is? in the question?

if it is simple x+y+z=12, then we have its solutions, so i guess its not a typo..

I am fairly sure it is not a typo ðŸ˜› If it is, then without knowing what it was supposed to be we can’t proceed.

regards

J

then can you plz explain what it is?

I don’t know! I didn’t post that question, someone else did. So unless he confirms if it is a typo or not there is nothing I can add.

regards

J

Sir, in the fishermen problem, what if I allot 7 fishes to each of 5. So that i need to take away 35-12=23 fishes from 5 men. So (23+4)C4=27C4?

Again, there would be cases where you are taking away more than 7 fish from a single person in that 27C4. In fact you will end up having more work, as when removing 23 fishes, there could be as many as 3 people who exceed the limit!

regards

J

Got it sir. Thank you very much. ðŸ™‚

How many 4 digit numbers are such that the sum of digits is equal to 15?

Is it going to be 17C3 – 8C3 – 3*7C3?

Something like that, yes.

regards

J

Thanks, wanted to confirm if the no. of cases we omit from 17C3 would be different for the 1st number and the rest.

sir in the fish question, taking that 7 as bound and writing the eqn has

7-a’+7-b’+7-c’+7-d’+7-e’=12

23=a’+b’+c’+d’+e’

but we know that the total fishes are itself 12. isn’t contradictory? I cant get hold of this. can u explaini

Two things:

First of all why would you do this? If I take the complement it is only to make life more simple. If the new number 23 is bigger than the original 12, then no point taking it.

Secondly, a’, b’ etc are not the fish, they are the shortage of fish. So if I tell you that in 3 subjects, out of 100 each you scored 110 marks. Then the marks you missed out on would be 300 – 110 = 190 right? i.e. if a + b + c = 110 then a’ + b’ + c’ = 190. That is what we are writing in the given problem.

regards

J

Yea. Thanks J :). Just another question J. If there are six cities which are to be connected by wires with the constraint that each city can have exactly incoming 2 wires from different cities in how many ways can one connect?

I am assuming that it is fine if we form two separate and disconnected closed loops. So then we have two options. (If not, only the first one)

Single ABCDEF loop which can be done in 5!/2 ways or Two loops ABC and DEF which can be done in 6C3 ways. (Do recheck the numbers as I did them in haste and may have made some mistake!)

regards

J

sir reg. 5!/2, the concept i used was that, a city can choose 2 out of 5 and so 6 cities doing the same will give me 5!*6 which gives me the same ans. but am i correct in doing so?

sorry sir that was 5c2*6

Not able to visualise your approach but if the answer matches then hopefully it must be fine!

regards

J

could you please elaborate second example 5 fisher men and 12 fishes i am getting some other answer.

Could you share your approach?

regards

J

Hello Sir..in the first example..a + b = 13 , but a cannot be 0, so a’ + b = 12, which gives 13 cases without any constraints..this is not considered in the example. I know something is wrong here as through manual calculation, the answer is correct (6). Can you please tell me what I am doing wrong

Please correct me if I am wrong..is the reason we are not considering a not equal to 0 in the initial case, because it will be neglected by taking out the values of b greater than 10?

That case is already eliminated because when a = 0, b would be 13 and we are removing all values of b > 9 ðŸ™‚ So it will be fine.

regards

J

Hello sir,

In Question 5, the fishing problem, according to the method new equation is a’ + b + c + d + e = 5 and there are 5 farmers so shouldn’t the answer should have 9C4 instead of 8C4 ?

got my mistake.

You have assumed no one can have 7 I think. No one can have *more than* 7, i.e. 8 or more, so we need to take a’ = a + 8.

regards

J

Sir, why arent we including the case where any two person can carry away with more than 7 fishes just like we are doing in the example post it, where any 2 person can sum upto more than 7? Does that means that will still effectively leave us with 5 to be distributed among rest 3, thus they;d be obvio below 7, that’s why we excluded those cases?

In first Question shouldn’t there be an extra requirement that a>=1 else it would not be a two digit number

Taken care of; since b cannot exceed 9 a has to be at least 4.

regards

J

oops missed that..Thanks a lot J……”Tum bahut mast kam karta hai”