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  • Question 1
    1 / -0

    Directions For Questions

     

    A tournament is arranged as per the following rules: in the beginning 16 teams are entered and divided in 2 groups of 8 teams each where the team in any group plays exactly once with all the teams in the same group. At the end of this round top four teams from each group advance to the next round in which two teams play each other and the losing teams goes out of the tournament. Then the winning teams play for semi finals and finally there is one final. The rules of the tournament are such that every match can result only in a win or a loss and not in a tie. 

    ...view full instructions

    The total number of matches played in the tournament is equal to

    Solution

    In the first round. Total no of matches = 2 × 8C2 = 2 × 28 = 56. Second round is knockout therefore total no.of match = 4 and 2 semi finals and one final thus total no. of matches in tournament = 63 

  • Question 2
    1 / -0

    Directions For Questions

     

    A tournament is arranged as per the following rules: in the beginning 16 teams are entered and divided in 2 groups of 8 teams each where the team in any group plays exactly once with all the teams in the same group. At the end of this round top four teams from each group advance to the next round in which two teams play each other and the losing teams goes out of the tournament. Then the winning teams play for semi finals and finally there is one final. The rules of the tournament are such that every match can result only in a win or a loss and not in a tie. 

    ...view full instructions

    Find the maximum number of matches that a team going out of the tournament in the first round itself can win.

    Solution

    Each team will play 7 matches and so any team can win any no .of matches between 0 to 7 0, 1, 2, 3, 4, 5, 6, 7. Four team will be selected (7, 6, 5, 4). Thus team which win only 3 matches will be out of the first round

  • Question 3
    1 / -0

    Directions For Questions

     

    A tournament is arranged as per the following rules: in the beginning 16 teams are entered and divided in 2 groups of 8 teams each where the team in any group plays exactly once with all the teams in the same group. At the end of this round top four teams from each group advance to the next round in which two teams play each other and the losing teams goes out of the tournament. Then the winning teams play for semi finals and finally there is one final. The rules of the tournament are such that every match can result only in a win or a loss and not in a tie. 

    ...view full instructions

    Which of the following statements about a team which has already qualified for the second round is true?

    Solution

    In second round it has to win one match. Then one in semifinal and one final. Total = 3

  • Question 4
    1 / -0

    Directions For Questions

    Number of ways of distributing 'n' different things into 'r' different groups is equal to 'rn'  when blank groups are taken into account and is equal to rn - rC1(r - 1)n  +  rC2 (r - 2)2 - .....+ (-1)r - 1  rCr - 1 when blank groups are not permitted. 

    ...view full instructions

    6 different balls have to be given to 3 children. Find the number of ways of distribution if each child gets at least one ball.

    Solution

    since each child received at least one ball. 

    It is equivalent to distribution   without blank groups .

    Required number of ways = 36  -  3C1 . 26 +3C2 = 540

     

  • Question 5
    1 / -0

    Directions For Questions

    Number of ways of distributing 'n' different things into 'r' different groups is equal to 'rn'  when blank groups are taken into account and is equal to rn - rC1(r - 1)n  +  rC2 (r - 2)2 - .....+ (-1)r - 1  rCr - 1 when blank groups are not permitted. 

    ...view full instructions

     A function is defined from a set A containing 6 elements to a set B containing 4 elements. In how many of these functions exactly one element of the set B is not an image

    Solution

    The element not an image can be chosen in 4C1 ways then 6 elements of A have to be

    distributed over 3 elements of B such that no elements of  B remains unassociated. 

    Thus the required number of functions 

           = 4C1[36  -  4C1 . 26+ 3C2 ] 
           = 4 × 540 
           = 2160

     

  • Question 6
    1 / -0

    Directions For Questions

    A cricket team of 11 players is to be selected from 8 batsmans, 6 bowlers, 4 all rounder and 2 wicket keepers. Answer the following question based on the information given.

    ...view full instructions

    Find the number of selections when at most 1 all rounder and 1 wicket keeper will play.

    Solution

    When 1 all rounder and 10 players from bowlers and batsman play the number of ways 

    = 4C1.14C10

    When 1 wicket keeper and 10 players from bowlers and batsman play the number of ways 

    = 2C1.14C10

    When 1 all rounder, 1 wicket keeper and 9 from batsman and bowlers play the number of ways

    = 4C1 . 2C1 . 14C9

    When all eleven players play from bowlers and batsman 

    Then the number of way = 14C11

    ∴ Total number of selections

    = 4C1 . 14C10 + 2C1 . 14C10 + 4C1 . 2C10 . 14C9 + 4C11.

     

  • Question 7
    1 / -0

    Directions For Questions

    A cricket team of 11 players is to be selected from 8 batsmans, 6 bowlers, 4 all rounder and 2 wicket keepers. Answer the following question based on the information given.

    ...view full instructions

    Find the number of selection when 2 particular batsmen don't want to play when a particular bowler will play.

    Solution

    If 2 batsman don't want to play then the rest of 10 players can be selected from 17 other

    players then number of selection = 17C10.

    If the particular bowler does not play then number of selection = 19C11

    If all the three don't play then number of selection = 17C11

    ∴ Total number of selection = 17C10 + 19C11 + 17C11.

     

  • Question 8
    1 / -0

    The largest integer 'n' such that 33! is divisible by 2n is

    Solution

    33!  = 1.2.3.4.5...........33

    = (2.4.6...........32)(1.3.5...........33)

    = 216 (1.2.3.4...........16)(1.3.5...........33)

    = 216 (2.4.6...........16)(1.3.5...........15)(1.3.5...........33)

    = 216.28 (1.2.3...........8)(1.3.5...........15)(1.3.5...........33)

    = 224 (2.4.6.8)(1.3.5.7)(1.3.5...........15)(1.3.5...........33)

    = 224.24 (1.2.3.4)(1.3.5.7)(1.3.5...........15)(1.3.5...........33)

    = 224.24 (1.2.3.4)(1.3.5.7)(1.3.5...........15)(1.3.5...........33)

    = 224.24.23 (1.3)(1.3.5.7)(1.3.5...........15)(1.3.5...........33)

    = 231(1.3)(1.3.5.7)(1.3.5...........15)(1.3.5...........33)

    Thus the maximum value of 'n' for which 33! is divisible by 2n is 31.

    The correct answer is: 31

     

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