Q-1) Find the number of triangles whose angular points are at the angular points of a given polygon of n sides, but none of whose sides are on the sides of the given polygon.
Solution:
The polygon has n sides – hence n angular points. We can choose a triangle from these n angular points in \(C(n,3)\) ways. Let us divide this set of triangles into 2 cases where:
a) At least one side of the triangle is a side of the polygon
b) None of the sides of the triangle lie among the sides of the polygon.
If, we consider one side of the polygon as a side of the triangle, then we can have a total of \(n-2\) triangles (Since there are remaining \(n-2\) angular points which could be the third vertex of the triangle). Thus, considering the entire polygon, we have a total of \(n(n-2)\) triangles which have at least one side of them as the edge of the polygon. However, in this case we have counted some triangles twice.
Let’s consider the polygon vertices to be \(A_1,A_2,A_3..,A_n\) in clockwise order. Consider that we are counting the side \(A_1A_2\) as a side of the triangle. In this case, we have considered the triangle \(A_1A_2A_3\) as a valid triangle. The same triangle has been considered when we considered \(A_2A_3\) as a side of the triangle. Thus the total number of unique triangles in this category are \(n(n-2)-n=n(n-3)\).
We’re interested in the number of triangles which fall in the second category and hence the number is \(C(n,3)-n(n-3)\).
Q-2) How many 7 digit numbers can you form whose sum of the digits is an even number?
Solution:
Let’s start by considering 10 successive 7 digit numbers, where the starting number ends with 0 and the last number ends with 9. In this set, the sum of the digits of each number is 1 more than the sum of the digits of the previous number. And thus, 5 of these numbers have an even sum of digits, while 5 of them have an odd sum of its digits. And thus, the parity of the sum of the digits can be determined by the last digit only.
Thus the first digit can be chosen in 9 ways, while the remaining 5 digits can each be chosen in 10 ways, while the last digit can again be chosen in 5 ways. This gives us the total number of ways as \(9\times 10^5\times 5=45\times 10^5\).
Q-3) In an examination, there are 3 papers each of 50 marks. The maximum marks for the fourth paper are 100. Find the number of ways in which a candidate can score 60% marks on the aggregate.
Solution:
Aggregate = 150+100=250
Thus, we need to find the number of ways in which the candidate can score 60% of 250=150.
Thus, using the concept of generating functions, we need to find the coefficient of \(x^{150}\) in \((x^0+x^1+x^2+…x^{50})^3(x^0+x^1+x^2+…x^{100})=(1-x^{51})^3(1-x^{101})(1-x)^{-4}\).
=coefficient of \(x^{150}\) in \((1-3x^{51}+3x^{102}-x^{153})(1-x^{101})(1+C(4,1)x+C(5,2)x^2+…)\)
=\(C(153,150)-3\times C(102,99)-C(52,49)+3\times C(51,48)=110556\).
Q-4) There are n straight lines in a plane, with \(n_r\) of them are parallel to the\(r\)th direction, where \(\sum\limits_{r=1}^k n_r=n\). Also, given that no three of them meet at a point. Show that the total number of points of intersection is \(\frac{1}{2}\left\{ n^2-\sum\limits_{r=1}^k n_r^2\right\}\).
Solution:
If we were told that no 2 lines are parallel and no three of them are concurrent, then we’d have a total of \(C(n,2)\) points of intersection. However, it’s given that there are \(k\) sets of parallel lines with no line in one set being parallel to any line in any other set. Thus the total number of intersection points is
Upon simplification, we arrive at the desired result.
Q-5) If all the permutations of the letters of the word SURITI are valid dictionary words, then what is the rank of the word SURITI in that dictionary among these permutations?
If we sort the letters of the word SURITI in alphabetical order, we get I,R,S,T,U.
The number of words beginning with I = 5!=120
The number of words beginning with R=\(\frac{5!}{2!}=60\).
After this, we have words beginning with S, thus number of words beginning with SI=4!=24, those with SR=\(\frac{4!}{2!}=12\), and similarly with ST is 12. With SUI, we have 3!=6 words, and then comes SURIIT and SURITI. Thus the rank of SURITI is 120+60+24+12+12+6+2=236.
Q-6) n different objects are arranged around a circle. In how many ways can you choose 3 of these objects such that no 2 of them are consecutive?
Solution:
Assume that the objects are marked as \(a_1,a_2,..,a_n\). Total number of ways of choosing 3 objects from these n objects is \(C(n,3)\). Now let’s consider the triplets where there are 2 or 3 consecutive elements. The number of such triplets with \(a_1\) are \(a_1a_2a_3,a_1a_2a_4,…,a_1a_2a_{n-1}\) – total of \(n-3\) triplets. (Ignore triplets like \(a_na_1a_2\), since we account for them when we consider our starting element of the triplet to be \(a_n\)).
Thus the total number of required triplets = \(C(n,3)-n(n-3)\).
Q-7) A is a set containing n elements. A subset P of A is chosen, and then A is reconstructed by replacing the elements of P. Now a subset Q is chosen from A. Find the number of ways of choosing P and Q such that \(|P\bigcap Q|=2\).
Solution:
If an element does not lie in the intersection of the 2 sets, then it can either be part of P, or it can be part of Q, or it can be neither a port of P nor Q. Thus there are 3 choices for an element if it is not part of the intersection of the 2 sets. Now choose any 2 elements of A and we put them in the 2 sets. This is done in \(C(n,2)\) ways. Now for the remaining n-2 elements there are 3 choices for each. Thus the total number of ways of choosing these sets = \(C(n,2)\times 3^{n-2}\).
Q-8) There are 2n guests at a party. The host and hostess have designated seats opposite to each other on the circular table, and there are 2 specified guests who cannot sit together. Find the number of ways in which the seating can be arranged.
Solution:
Let’s call the specified guests P and Q. Consider the 2n seats across the circular table to be \(a_1,a_2,…,a_{2n}\).
Consider the above figure. If we put P at \(a_1\) and Q at any position other than \(a_2\), then the remaining 2n-2 guests can be arranged in \((2n-2)!\) ways. Thus, altogether there will be \((2n-2)(2n-2)!\) arrangements if P is seated at \(a_1\). The same goes if P sits at any of \(a_n,a_{n+1},a_{2n}\), and hence for all of these 4 seats including \(a_1\), we have \(4\times (2n-2)\times (2n-2)!\) ways.
Now, let’s say if P is at \(a_2\), in that case there will be (2n-3) positions for Q, implying that there will be altogether \((2n-3)(2n-2)!\) arrangements if P is at \(a_2\). And same if P does not sit at any of the seats \(a_1,a_n,a_{n+1},a_{2n}\). That gives a total of \((2n-4)\times (2n-3)\times (2n-2)!\) arrangements of the guests.
Adding the first and second case, we have the total number of ways as
\(4(2n-2)(2n-2)!+(2n-4)(2n-3)(2n-2)!=(4n^2-6n+4)(2n-2)!\).
Q-9) There are n straight lines in a plane. Exactly \(p\) points out of these are collinear.
a) Find the number of straight lines that can be formed by joining them
b) Find the number of triangles that can be formed by joining them
Solution:
a) A straight line can be formed by joining any 2 points. The total number of such straight lines form n given points = \(C(n,2)\). However, note that joining any number of the collinear points are going to give only one straight line. Thus the number of straight lines is \(C(n,2)-C(p,2)+1\).
b) We can form a triangle by choosing any 3 of the given set of points, however we can’t have any combination of 3 points from the given set of p points. Thus the total number of triangles – \(C(n,3)-C(p,3)\).
Q-10) Consider that we have to make a 4 digit number from the first 7 digits. How many of these numbers will be divisible by 25?
Solution:
A number is divisible by 25 if the last 2 digits of the number is divisible by 25. Thus the last 2 digits can be – 25,75.
Thus we choose any 2 digits from the remaining 5 digits. The number of ways will be \(^5P_2\times 2\).