if set A has m elements , and set B has n elements ..then based on the student and classroom analysis ….no of relationships defined from set A to set B should be n^m …??
Right you are. But if you think about it, whether we take m^n or n^m does not matter here – the point is that we need how many pairs are there such that something^something = 4096 š There are only 6 pairs which give 4096…(the listing at the end is in the correct order of n^m)
There are p committees in a class (where pā„5), each consisting of q members (where qā„6).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?
and there are 4 options.
a)(p-1)(q-1) b) pq-C(q,2),pq c) pq-C(p,2),pq d)pq-p,pq-1
This is the original CAT question on which the one you’ve mentioned is based! š (The answer to your question would be option (c) as I am sure you will be able to work out now)
Find max total number of students (p*q). Find max possible “common” students. Subtract. In this case there are p groups and any two groups can have a student in common. So there would be at max pC2 common students. So minimum = pq – pC2.
In the original CAT question that you referred to for Sanya’s question, the number of players counted twice need not necessarily be n. It could be anywhere between 1 & n because there one player could be common between multiple pairs of groups. How can we have this assumption that all the common players are distinct?
And in Sanya’s question, the maximum possible repeated counting of members will be pC2 only if p+1>=q. If p+1q and trying to map the members. So, how can we assume that p+1>=q?
Sir,
In the original CAT question that you referred to for Sanya’s question, the number of players counted twice need not necessarily be n. It could be anywhere between 1 & n because there one player could be common between multiple pairs of groups. How can we have this assumption that all the common players are distinct?
And in Sanya’s question, the maximum possible repeated counting of members will be pC2 only if pq+1, then that value will be less than pC2. This can be proven by taking any small value of p, q such that p>q+1 and trying to map the members. So, how can we assume that p<=q+1?
Because we need the extreme cases, maximum or minimum. All in-between values could also be taken! For your second question, I would agree with you that it should be specified. However, whatever the numbers involved, it cannot go less that the third answer so that still provides a “hard minimum”.
regards
J
Shouldn’t it be pq – p (for p groups having one student in common)?
I have a problem in the way the ordered pair (m,n) has been calculated. As far as I know, The total number of relations from A to B is given by 2^pq , where n(A)=p and n(B)=q. Using this approach the product of p and q comes out to be 12.
Then,{1,12}{12,1}{4,3}{,3,4}{6,2}{2,6} should be the ordered pairs.
Though the answer,6, remains the same.
Agreed Vaibhav, if you know the formula š However, I try to avoid a formulaic approach and look more at understanding-based solutions whenever possible, so that readers who are not that comfortable in maths, or don’t have extensive grounding in it (which sums up the majority!) should still find that the problem is solvable within their comfort level. A lot of things in PnC particularly have a direct formula, but one that is hard to understand! Thanks for your comment…
CAT-holics 
if set A has m elements , and set B has n elements ..then based on the student and classroom analysis ….no of relationships defined from set A to set B should be n^m …??
Right you are. But if you think about it, whether we take m^n or n^m does not matter here – the point is that we need how many pairs are there such that something^something = 4096 š There are only 6 pairs which give 4096…(the listing at the end is in the correct order of n^m)
Sir,
Can you help me with this question please :
There are p committees in a class (where pā„5), each consisting of q members (where qā„6).No two committees are allowed to have more than 1 student in common. What is the minimum and maximum number of students possible?
and there are 4 options.
a)(p-1)(q-1) b) pq-C(q,2),pq c) pq-C(p,2),pq d)pq-p,pq-1
Sanya,
Try the last question in this and see whether it helps.
https://crackthecat.wordpress.com/2013/05/30/fpc-and-pnc-cat-scan-1/
This is the original CAT question on which the one you’ve mentioned is based! š (The answer to your question would be option (c) as I am sure you will be able to work out now)
regards
J
I tried to do this question using that method but I did not get the right answer. What should be the right approach??
Find max total number of students (p*q). Find max possible “common” students. Subtract. In this case there are p groups and any two groups can have a student in common. So there would be at max pC2 common students. So minimum = pq – pC2.
regards
J
Sir,
In the original CAT question that you referred to for Sanya’s question, the number of players counted twice need not necessarily be n. It could be anywhere between 1 & n because there one player could be common between multiple pairs of groups. How can we have this assumption that all the common players are distinct?
And in Sanya’s question, the maximum possible repeated counting of members will be pC2 only if p+1>=q. If p+1
Sir,
In the original CAT question that you referred to for Sanya’s question, the number of players counted twice need not necessarily be n. It could be anywhere between 1 & n because there one player could be common between multiple pairs of groups. How can we have this assumption that all the common players are distinct?
And in Sanya’s question, the maximum possible repeated counting of members will be pC2 only if pq+1, then that value will be less than pC2. This can be proven by taking any small value of p, q such that p>q+1 and trying to map the members. So, how can we assume that p<=q+1?
Because we need the extreme cases, maximum or minimum. All in-between values could also be taken! For your second question, I would agree with you that it should be specified. However, whatever the numbers involved, it cannot go less that the third answer so that still provides a “hard minimum”.
regards
J
Shouldn’t it be pq – p (for p groups having one student in common)?
Note that p is smaller than pC2. We want to subtract as many as possible, to get the minimum number….
regards
J
I have a problem in the way the ordered pair (m,n) has been calculated. As far as I know, The total number of relations from A to B is given by 2^pq , where n(A)=p and n(B)=q. Using this approach the product of p and q comes out to be 12.
Then,{1,12}{12,1}{4,3}{,3,4}{6,2}{2,6} should be the ordered pairs.
Though the answer,6, remains the same.
Agreed Vaibhav, if you know the formula š However, I try to avoid a formulaic approach and look more at understanding-based solutions whenever possible, so that readers who are not that comfortable in maths, or don’t have extensive grounding in it (which sums up the majority!) should still find that the problem is solvable within their comfort level. A lot of things in PnC particularly have a direct formula, but one that is hard to understand! Thanks for your comment…
regards
J
Sir, in the last (m,n) pairs, I guess the alphabets are interchanged. It works for (n,m) over there, but maybe a typo did it (m,n) š
Yes, it should have been (n, m). Typo on my part, sorry!
regards
J