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6.
There are n teams participating in a football championship. Every two teams played one match with each other. There were 171 matches on the whole. What is the value of n?
  • A.
    21
  • C.
    17
  • B.
    19
  • D.
    15
  • Answer & Explanation
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Answer : [B]
Explanation :
Total number of matches played = nC2
Given, nC2 = 171
n(n - 1)
2
 = 171
  n2 - n - 342 = 0  
    (n +18) (n-19) = 0    
  n = 19 (   n cannot be neative)
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7.
How many words can be formed from the letters of the word AKHILESH, so that the vowels are always together?
  • A.
    2160
  • C.
    2140
  • B.
    3620
  • D.
    360
  • Answer & Explanation
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Answer : [A]
Explanation :
In the given word, AKHILESH, we treat the vowels AIE as one letter.
Thus, we have KHLSH (AIE)
This group has 6 letters of which H occurs 2 times and other letters are different.
Number of ways of arranging these letters =
6!
2!
 = 360  
Now 3 vowels can be arranged among themselves in 3! = 6 ways
    Total number of ways = 360 x 6 = 2160  
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8.
5 men and 5 women sit around a circle table, the men and women alternately. In how many different ways can the seating arrangement be made?
  • A.
    2800
  • C.
    10!
  • B.
    2880
  • D.
    9!
  • Answer & Explanation
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Answer : [B]
Explanation :
5 men can be arranged in a circular table in 4! Ways = 24 ways
There are 5 seats available for the five women
They can be arranged in 5! Ways
The number of ways in which the men and women can be seated = 5! X 4! Ways = 2880 ways.
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9.
There are 6 boxes labeled A, B, C, D, E and F. In each box a black or a white ball is to be put in such a manner that at least one box contains a black ball and the boxes containing black balls are labeled with consecutive letters. The total number of ways in which  this can be done is
  • A.
    21
  • C.
    20
  • B.
    19
  • D.
    26
  • Answer & Explanation
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Answer : [A]
Explanation :
The boxes containing the black balls can be as follows:
  (i)   One box containing a black ball (i.e) A, B, C, D, E or F
       
  (ii)   Two boxes containing black balls (i.e) AB, BC, CD, DE or EF
       
  (iii)   Three boxes containing black balls (i.e) ABC, BCD, CDE, DEF
       
  (iv)   Four boxes containing black balls (i.e) ABCD, BCDE, CDEF
       
  (v)   Five boxes containing black balls (i.e.) ABCDE, BCDEF
       
  (vi)   Six boxes containing black balls ABCDE.
       
    Total number of ways = 6 + 5 + 4 + 3 + 2 + 1 = 21 ways
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10.
In how many ways can 7 MBA and 5 MCA students be seated in a row so that no two of the MCA students may sit together?
  • A.
    7! 8!
  • C.
    		 7! 8!  
    3!  
  • B.
    		 7! 8!  
    2!  
  • D.
    56
  • Answer & Explanation
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Answer : [C]
Explanation :
There are no restriction on MBA Students
    We first fix the position of 7 MBA students as follows:
|x| MBA |x| MBA |x| MBA |x| MBA |x| MBA |x| MBA |x| MBA |x|
Now, if MCA students sit at the places (including the two ends) indicated by |x|,
then no two of the 5 MCA students will come together.
There 8 places are filled by 5 MCA students in 8P5 ways.
The 7 MBA students can sit in a row in 7! Ways
 
  Required number of ways = 8P5 x 7! =
8!
3!
 x 7!  
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