Company BELIANCE hosted a party for 8 members of Company AXIAL. In the party no member of AXIAL had interacted with more than three members of BELIANCE. Out of all the members of BELIANCE, three members – each interacted with four members of AXIAL and the remaining members – each interacted with two members of AXIAL. The greatest possible number of members of company BELIANCE in the party is:
Answer: A The important constraint here is that in the party, no member of AXIAL had interacted with more than three members of BELIANCE. Given that there are 8 members of company AXIAL and three members of company BELIANCE interacted with four members of AXIAL. Therefore, the maximum possible number of members of company BELIANCE in the party will be 3+ 4C2 (Since, each of the remaining members of the company BELIANCE haveinteracted with two members of AXIAL) = 9.
Q. No. 2:
If F(x, n) be the number of ways of distributing “x” toys to “n” children so that each child receives at the most 2 toys then F(4, 3) = _______?
Answer: B We have to find the number of ways in which 4 toys can be distributed to 3 children so that each child receives at the most 2 toys. There are two possible cases: Case 1: Two of them receive 2 toys each and one of them doesn‟t get any toy. There are 3 possible ways to distribute the toys in this case i.e., the three possible ways of selecting the child who will not get any toy. Case 2: Two of them receive 1 toy each and one of them receives 2 toys. Again there are 3 possible ways to distribute the toys in this case i.e., the three possible ways of selecting the child who will get 2 toys. There are a total of 6 possible ways.
Q. No. 3:
Suppose you have a currency, named Miso, in three denominations: 1 Miso, 10 Misos and 50 Misos. In how many ways can you pay a bill of 107 Misos?
Answer: C Let the number of currency 1 Miso, 10 Misos and 50 Misos be x, y and z respectively. x + 10y + 50z = 107 Now the possible values of z could be 0, 1 and 2. For z = 0: x + 10y = 107 Number of integral pairs of values of x and y that satisfy the equation x + 10y = 107 will be 11. These values of x and y in that order are (7, 10); (17, 9); (27, 8)…(107, 0). For z = 1: x + 10y = 57 Number of integral pairs of values of x and y that satisfy the equation x + 10y = 57 will be 6. These values of x and y in that order are (7, 5); (17, 4); (27,3); (37, 2); (47, 1) and (57, 0). For z = 2: x + 10y = 7 There is only one integer value of x and y that satisfies the equation x + 10y = 7 in that order is (7, 0). Therefore total number of ways in which you can pay a bill of 107 Misos = 11 + 6 + 1 = 18
Q. No. 4:
The number of ways of arranging n students in a row such that no two boys sit together and no two girls sit together is m(m >100). If one more students is added, then number of ways of arranging as above increases by 200%. The value of n is
Answer: D If n is even, then the number of boys should be equal to number of girls, let each be 'a' => n =2a Then the number of arrangements = 2*a!*a! If one more students is added, then number of arrangements = a!* (a+1)! But this is 200% more than the earlier => 3(2*a!*a!) = a!*(a+1)! => a+1 = 6 and a =5.=> n=10. But if n is odd, then number of arrangements = a!(a+1)! where, n = 2a+1 When one student is included, number of arrangements = 2(a+1)!(a+1)! => By the given condition, 2(a+1) = 3, which is not possible.
Q. No. 5:
Some boys are standing on a circle at distinct points. Each possible pair of persons, who are not adjacent, sing a 3 minute song, one pair after another. The total time taken by all the pairs to sing is 1 hour. Find the number of boys?
Answer: C Each boy would pair with n-3 other boys Number of possible pairs = n(n-3)/2 n(n-3)/2 = 60/3 = 20 => n(n-3)= 40 => n = 8
Q. No. 6:
There are 10 seats around a circular table. If 8 men and 2 women have to seated around a circular table, such that no two women have to be separated by at least one man. If P and Q denote the respective number of ways of seating these people around a table when seats are numbered and unnumbered, then P/Q equals
Answer: C Initially we look at the general case of the seats not numbered. The total number of cases of arranging 8 men and 2 women, so that women are together => 8!2! The number of cases where in the women are not together => 9! - (8!2!) = Q Now, when the seats are numbered, it cab be considered to a linear arrangement and the number of ways of arranging the group such that no two women are together is => 10! - (9!2!). But the arrangements where in the women occupy the first and the tenth chairs are not favourable as when the chairs which are assumed to be arranged in a row are arranged in a circle, the two women would be sitting next to each other. The number of ways the women can occupy the first and the tenth position = 8!2! The value of P = 10! - (9!2!) - (8!2!) Thus P/Q = 10/1