Demystifying Permutation & Combination - Rajesh Balasubramanian - CAT 100th Percentile - CAT 2011, 2012 and 2014 (Part 5/7)


  • Director, 2IIM Online CAT Preparation | IIT Madras | IIM Bangalore | CAT 100th percentile - CAT 2011, 2012 and 2014.


    Diagonals of a square ABCD of side 35 cms intersect at O. A rhombus of perimeter 52 exists such that P lies on AO, Q on BO, R on CO and S on DO. What is the maximum number of circles with integer radii that can be drawn with center O such that the rhombus is inside the circle and the circle is inside the square?

    We are trying to draw circles that are inside the square. So, radius should be less than
    35/2 = 17.5. Or, maximum integer radius should be 17.
    We are trying to draw circles such that the rhombus should be inside the circle. So, the diameter of the circle should be greater than the longest diagonal of the rhombus. Or, the longer diagonal of the rhombus should be as short as possible. For a rhombus of the given perimeter, this is possible only if it is a square. A square of perimeter 52cm will have diagonals of length 13√2.
    Diameter of circle > 13√2
    Or radius > 13/√2
    Or, radius > 10
    Possible values of radius = 10, 11, 12, 13, 14, 15, 16, 17 = 8 different values.

    The diagonals of Hexagon intersect at n distinct points inside the hexagon. What is the maximum value n can take?

    In any hexagon, there are n (n - 3)/2 = 9 diagonals.
    First let us draw the diagonals and try to visualise this diagram

    0_1487674891712_122.png

    There are six ‘short’ diagonals AC, AE, CE, BD, BF, and DF. These intersect with other diagonals at 3 points each.
    There are 3 long diagonals – AD, BE and CF. These intersect with other diagonals at 4 points each.
    Note that the ‘short’ diagonals need not be shorter than the ‘long’ diagonal.
    So, the total number of points of intersection should be 6 × 3 + 3 × 4 = 30.
    But in this case, we would count every point of intersection twice. So, number of points of intersection would be exactly half of this = 30/2 = 15 points.

    Consider a circle of radius 6 cms. What is the maximum number of chords of length 6 cms that can be drawn in the circle such that no two chords intersect or have points of contact?

    First let us think about the angle subtended at the center by this chord. In a circle of radius 6cms, a chord of length 6 cms subtends an angle of 60° at the center. This chord along with the two radii forms an equilateral triangle.
    So, we can place six such equilateral triangles at the center to account for 360°. This would form a regular hexagon. But in this scenario, the chords would have points of contact. So, the maximum number of chords that can be drawn such that there are no points of contact is 5.
    More generally, if a chord makes an angle q at the center, then we can draw 360/q such chords around the circle. The maximum number of chords that can be drawn such that they do not touch each other = (360/q) – 1.
    If 360/q is not an integer, then the maximum number of chords that can be drawn such that they do not touch each other = [360/q], where [x] is the greatest integer less than or equal to x.

    x (x – 3) (x +2 ) < 200, and x is an integer such that |x| < 20, how many different values can n take?

    Let us start with a trial and error. The expression is zero for x = 0, x = 3 and x = –2
    x = 3, the above value = 0
    x = 4, the above value would be 4 × 1 × 6 = 24
    x = 5, the above value would be 5 × 2 × 7 = 70
    x = 6, the above value would be 6 × 3 × 8 = 144
    x = 7, the above value would be 7 × 4 × 9 > 200
    So, the equation holds good for x = 3, 4, 5, 6.
    For x = 2 and 1, the above value is negative.
    So, the above inequality holds good for x = 6, 5, 4, 3, 2, 1, 0.
    For, x = –1, the value would be –1 × –4 × 1 = 4.
    For, x = –2, the value would be 0.
    So, this works for x = 6, 5, 4, 3, 2, 1, 0, –1, –2.
    For, x = –3, the expression is negative, so holds good. For all negative values < –3, this holds good.
    The smallest value x can take is –19.
    So, the above inequality it holds good for –19, –18, –17…..–1, 0, 1 ……6, a total of 26 values.

    In how many ways can we pick three cards from a card pack such that they form a sequence of consecutive cards, not all cards belong to the same suit, and nor do all cards belong to distinct suits? Consider Ace to be the card following King in each suit. So, Ace can be taken to precede ‘2’ and succeed ‘King’. So, QKA would be a sequence, so would be A23. However, KA2 is not a sequence.

    First let us see how many sequences of 3 we can form. We can have A23, 234…..JQK, QKA – a total of 12 sets of 3.
    If cards should not be of the same suit, and nor should all three be of different suits, then we should have two cards from one suit and one from another.
    So, cards should be from two suits. The two suits can be selected in 4C2 ways. Now, from these two suits, one suit should have two cards. The suit that has two cards can be selected in 2C1 ways. Now, out of the three cards, the two cards that have to be from the suit that repeats can be selected in 3C2 ways.
    So, total number of possibilities = 12 × 4C2 × 2C1 × 3C2 = 12 × 6 × 2 × 3 = 432.

    Given that |k| < 15, how many integer values can k take if the equation x^2 – 6|x| + k = 0 has exactly 2 real roots?

    The equation can be rewritten as |x|^2 – 6|x| + k = 0. This is a quadratic in |x|. This can have 2 real roots, 1 real root or 0 real roots.
    If we have |x| = positive value, we have two possible values for x.
    If we have |x| = negative value, we have no possible values for x.
    If we have |x| = 0, we would have 1 possible value for x.
    So, for the equation to have 2 values of x, we should have 1 positive root for |x|.
    Scenario I: |x|^2 – 6|x| + k has exactly one real root (and that root is positive). b^2 – 4ac = 0 => k = 9. If k = 9, |x| = 3, x can be 3 or –3
    Scenario II: |x|^2 – 6|x| + k has two real roots and exactly one of them is positive. This tells us that the product of the roots is negative. => k has to be negative. |K| has to be less than 15.
    = > k can take values –14, –13, –12, ….–1 : 14 different values
    Total possibilities = –14, –13, –12, ….–1 and k = 9; 15 different values

    In how many ways can 6 boys be accommodated in 4 rooms such that no room is empty and all boys are accommodated?

    No room is empty, so the boys can be seated as 1113 in some order or 1122 in some order.
    Scenario I: 1113. We can do this as a two–step process.

    Step I: Select the three boys – 6C3.
    Step II: Put the 4 groups in 4 rooms – 4! ways
    Total number of ways = 20 × 4!
    Scenario I: 1122. This is slightly tricky. So let us approach this slightly differently.
    Step I: Let us select the 4 people who are going to be broken as 2 + 2; this can be done in 6C4 ways. Now, these 6C4 groups of 4 can be broken into 2 groups of two each in 4C2 / 2 ways. So, the total number of ways of getting 2 groups of 2 is 6C4 × 4C2/2 = 15 × 6/2 = 45 ways
    Step II: Now, we need to place 2, 2, 1, 1 in four different groups. This can be done in 4! ways.
    The total number of ways = 20 × 4! + 45 × 4! = 4! (20 + 45) = 24 × 65 = 1560 ways.

    In how many ways can 4 boys and 4 girls be made to sit around a circular table if no two boys sit adjacent to each other?

    No two boys sit next to each other => Boys and girls must alternate. As they are seated around a circular table, there is no other possibility.
    Now, 4 boys and 4 girls need to be seated around a circular table such that they alternate. Again, let us do this in two steps.
    Step I: Let 4 boys occupy seats around a circle. This can be done in 3! ways.
    Step II: Let 4 girls take the 4 seats between the boys. This can be done in 4! ways.
    Note that when the girls go to occupy seats around the table, the idea of the circular arrangement is gone. Girls occupy seats between the boys. The seats are defined as seat between B1 & B2, B2 & B3, B3 & B4 or B4 & B1. So there are 4! ways of doing this.
    Total number of ways = 3! × 4! = 6 × 24 = 144

    John extracts three letters from the word ‘ACCEDE’ and makes words out of them, how many different words can he generate?

    John can extract three distinct letters or, 2 of one kind and one different.
    Scenario I: Three distinct letters
    Step I: Some 3 of the 4 letters A, C, D, E can be selected. 4C3
    Step II: This can be rearranged in 3! ways.
    Total number of ways = 4C3 × 3! = 4 × 6 = 24.
    Scenario II:
    Step I
    (i) 2 C’s and one of A, D or E or
    (ii) 2 E’s and one of A, C or D. 6 possibilities totally
    Step II: This can be rearranged in 3!/2! ways
    Total number of ways = 6 × 3 = 18
    24 + 18 = 42

    Joseph extracts three letters from the word ‘RENEGED’ and makes words out of these 3 letters. How many such words can he generate?

    Joseph can extract three distinct letters or, 2 of one kind and one different, or all 3 being the same letter.
    Scenario I: Three distinct letters
    Step I: Any 3 of the 5 letters R, E, N, G, D. 5C3
    Step II: This can be rearranged in 3! = 6 ways.
    Total number of ways = 5C3 × 3! = 10 × 6 = 60
    Scenario II:
    Step I: 2 E’s and one of R, G, D or N. 4 ways of selecting one of the other 4 letters.
    Step II: This can be rearranged in 3!/2! = 3 ways
    Total number of ways = 4 × 3 = 12
    Scenario III: All three being the same letter. All three can be ‘E’. There is only one word.
    60 + 12 + 1 = 73


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