1. Find the no. of three element set of positive integers (a,b,c) satisfying the abc = 2310
2. Find the no. of distinct throws that can be made with 'n' 6 faced fair dices which are indistinguishable among them?
NOTE: Please let me know In what sense do we need to interpret the word distinct here?
3. If r,s,t are prime nos. and p,q are the positive integers such that their LCM of p,q is r2t4s2, the no. of ordered pair of (p,q) is - ?
4. A rectangle of sides 2m-1 and 2n-1 is divided into squares of UNIT length by drawing parallel lines, then the no. of rectangles possible with ODD side lengths is - ?
5. Given a positive Integer n, then find the number of qudrapules (a,b,c,d) such that 0\leq a\leq b\leq c\leq d\leq n
716. How many vectors (a_{1},a_{2},........,a_{k}) with integral a_{i} \epsilon \left\{1,2,.......,n \right\} are there satisfying 1\leq a_{1}\leq a_{2}\leq a_{3}....\leq a_{k}\leq n?
21can u please post the answers by hiding.
that makes easier to verify
71Actually, That is the skill which I want to develop too.. Like having 100% confidence in one's solution. Cause in combinatorics I am having great problem related to that. I solve any problem but can't guarantee it to be right (although in most cases it is). What should I do?
I guarantee that the questions are right, but not the answers cause a friend of mine gave me this questions. (the first two)
1. 40 3. 225
39q3)
www.fiitjee.com/down/sol/maths06.pdf
or
http://www.scribd.com/IITJEE-Solved-Mathematics-2006/d/2227207
391)
2310= 1x2x3x5x7x11
try seeing the question as number of ways in which 3 pairs can be selected from 6 items
30Q4). We need to select odd side lengths.
So, possibilities : (2m-1)(2n-1), (2m-1)(2n-3), (2m-1)(2n-7) ..
Total no of ways = (2m-1){(2n-1) + (2n-3) + (2n-5) + (2n-7) + ....} + (2m-3){(2n-1) + (2n-3) + (2n-5) + (2n-7) + ....} + ....
= [(2m-1) + (2m-3) + (2m-5) + ...][(2n-1) + (2n-3) + (2n-5) + ...]
= [m/2(2m-1 + 1)][n/2(2n-1 + 1)]
= m2n2
71@phophet Sir,
Atleast show me how are you proceeding sir? Please
@Ashish, Thanks! Please look the other ones too!!
3412310 = 2 X 3 X 5 X 7 X 11
A non-trivial factor is a factor other than 1.
If there is only one non-trivial factor, then the triplet is (1,1,2310)
If there is one trivial factor, then we are looking for ways to factorise 2310 into two factors
If no trivial factor, then 2310 is to be factorised into three factors.
i.e. the numbers 2,3,5,7,11 are to be divided into two and three groups respectively. Here the concept of Stirling numbers of the second kind comes into play. Check out http://en.wikipedia.org/wiki/Stirling_numbers_of_the_second_kind
Hence, the number of triplets is 1+ S(5,2)+S(5,3) = 41
[The link gives the stirling numbers for some cases]
71Sir, One doubt. The solutions given by Stirling nos. are non distinct i.e., 1+2+3 = 3+2+1 . But is this the case here too? Cause we are assigning values to three different variables viz., a,b,c.
So If we have {1,1,2310}, then we can have {1,2310,1} and {2310,1,1} too ?
341That is why the have said sets. So, any permutation is considered the same