TCS
Company
Numerical Ability
Permutation and Combination
An organization has 4 committees. Only 3 persons are members of all four committees, but every pair of committees has 4 members in common. What is the LEAST possible number of the members on any one committee?
Read Solution (Total 14)
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- 6 least possible no. of members in any one committee.
C1 C2 C3 C4
x x x x
y y y y
z z z z
w w v t
r v t u
s u s r
now every two committee has 4 members in common.
ans : 6 - 10 years agoHelpfull: Yes(26) No(7)
- 4 with help of set diagram
- 10 years agoHelpfull: Yes(10) No(1)
- let w,x,y,z are four committees....A,B,C are common in all committee...
w x y z
__________
A A A A
B B B B
C C C C
D D _ D
E E E _
F _ F F
__________
pairs of the commiittee -->AB,AC,AD,BC,BD,CD
IN AB-->A,B,C,D are common
IN AC-->A,B,C,E are common
IN AD-->A,B,C,F are common
IN BC-->A,B,C,E are common
IN BD-->A,B,C,D are common
IN CD-->A,B,C,F are common
SO,the LEAST possible number of the members on any one committee is= 5 (ans.) - 10 years agoHelpfull: Yes(9) No(1)
- 4 with the help of ven diagram
- 10 years agoHelpfull: Yes(5) No(2)
- the least possible no of member of members on any one committee = 4 .. option a)
In all 4 committees, say W,X,Y,Z, 3 persons say A,B and c are common.
D is common between W and X.
E is common between X and Y.
F is common between Y and Z.
G is common between Z and W.
so W committee has ABCDG. X committee has ABCDE. Y committee has ABCEF.
Z committee has ABCFG.
- 10 years agoHelpfull: Yes(3) No(4)
- 1 2 3 4
A A A A
B B B B
C C C C
e e f d
d f d
Here,1 2 3 4 are the committees,and A B C are the members common to all the committees,A B C e is common to 1 2,A B C f is comm to 2 3,A B C d is comm to 3 4, A B C d is comm to 4 3 so ans is the 4th committee which has least mem 4 - 10 years agoHelpfull: Yes(2) No(1)
- (ABC)DGI
(ABC)DFH
(ABC)EGH
(ABC)EFI - 10 years agoHelpfull: Yes(2) No(0)
- 8 with help of set diagram
- 10 years agoHelpfull: Yes(1) No(8)
- 6 people using venn diagram
- 10 years agoHelpfull: Yes(1) No(1)
- let w,x,y,z are four committees....A,B,C are common in all committee...
w x y z
__________
A A A A
B B B B
C C C C
D D _ D
E E E _
F _ F F
__________
pairs of the commiittee -->AB,AC,AD,BC,BD,CD
IN WX-->A,B,C,D are common
IN WY-->A,B,C,E are common
IN WZ-->A,B,C,F are common
IN XY-->A,B,C,E are common
IN XZ-->A,B,C,D are common
IN YZ-->A,B,C,F are common
SO,the LEAST possible number of the members on any one committee is= 5 (ans.) - 10 years agoHelpfull: Yes(1) No(0)
- 0 is the answer cause here 3 members are there for all 4 committees
- 10 years agoHelpfull: Yes(0) No(6)
- Let the committies be 1,2,3 and 4 then,
A,B,C are common to all the four commities(since given in the que that only 3 persons are the members of all the four committies)
now, D is common to 1&2
E is common to 2&3
F is common to 3&1
so, the least possible number are 4
ANSWER: 4
- 10 years agoHelpfull: Yes(0) No(1)
- let the committees be 1234
1 2 3 4
a a a a
b b b b
c c c c (since 3 mem common in all four committee)
d d
e e
f f
g g
i i
h h (each pair of committee has 4 members in common)
therefore least possible number of mem in any group is 6
- 10 years agoHelpfull: Yes(0) No(0)
- 6 is the answer.
Lets us assume A,B,C,D are 4 committees.
now using Venn diagram ,
A ∩ B ∩ C ∩ D =3
Now every pair must have 4 members in common
possible pairs are(each containing +1 member since there're already 3 common members) :
A ∩ B - 1
A ∩ C - 1
A ∩ D - 1
B ∩ C - 1
B ∩ D - 1
C ∩ D - 1
Now if we take A alone there are
3+ 1 (in A ∩ B)+1 (in A ∩ C ) + 1 (in A ∩ D)= 6
Similarly if u check for all , there will be 6 members in each.(minimum possibility)
For even more clear understanding , apply this in a Venn diagram. - 6 years agoHelpfull: Yes(0) No(0)
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