1. Good post? |

1.Eighteen guests have to be seated, half on each side of a long table,four particular guests desire to sit on one particular side and three others on the other side.Determine the number of ways in which the sitting arrangements can be made

(A) 9! 11P5 (B) 9! 11P9 (C)9! 11P5 (D) none

2.The number of ways in which a team of seven players can be selected from 22 players always including 2 of them and excluding 4 of them is

(A) 16C11 (B) 11C4 (C)16C9 (D)none

3.The number of unsuccessful attempts that can be made by a theif to open a NUMBER lock having 4 rings in which each ring contains 4 numbers is

(A) 620 (B) 621 (C)642 (D)none

4.The numbers of numbers which are not divisible by 5 can be formed by the digits 0,2,4,5 without repitition is
(A) 20 (B) 21 (C)22 (D)23

5.The number of arrangements that can be made out of the letters of word "MISSISSIPI" so thatall the S's come together are
(A) 210 (B)211 (C) 212 (D) 213

5.The number of arrangements that can be made out of the letters of word "MISSISSIPI" so that all the S's come together and the I's not come together is
(A) 185 (B)184 (C) 180 (D) 186

6.In how many ways can five men sit around a table so that all shall have the same neighbours in any two arrangements is
(A) 11 (B) 10 (C) 12 (D) none

7.There are 4 doors to a lecture roonm.The number of ways that a student enter the room and leave it by different door
(A) 12 (B)13 (C)10 (D) none

8.15 passengers are to travel by a double decker bus,which can accomodate 5 in the upper desk and 10 in the lower desk.The number of ways that these passengers are distributed is
(A)3104 (B)3105 (C)3106 (D)none

9.A father takes 8 children,3 at a time to the zoo, as often as he can take without the same three together more than once, the number of times each child go is equal to the number of times
can he go is
(A) 2155 (B)2156 (C)2153 (D)none

2. Good post? |
can ne1 provide us the problems..with correct wording and the official answer.

9.A father takes 8 children,3 at a time to the zoo, as often as he can take without the same three together more than once, the number of times each child go is equal to the number of times
can he go is
(A) 2155 (B)2156 (C)2153 (D)none

8 C 3 = 56...number of times the father goes
7 C 2 = 21...number of times each child go

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