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a+n+p=12
a,n,p are integers from 1 to 7.Find no. of possible solutions?
Read Solution (Total 16)
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- Total no. of solutions - 37
combinations such as -
(1,4,7)(1,5,6)(1,6,5)(1,7,4) - 4
(2,3,7)(2,4,6)(2,5,5)(2,6,4)(2,7,3) - 5
(3,2,7)(3,3,6)(3,4,5)(3,5,4)(3,6,3)(3,7,2) - 6
(4,1,7)(4,2,6)(4,3,5)(4,4,4)(4,5,3)(4,6,2)(4,7,1) - 7
(5,1,6)(5,2,5)(5,3,4)(5,4,3)(5,5,2)(5,6,1)- 6
(6,1,5)(6,2,4)(6,3,3)(6,4,2)(6,5,1) - 5
(7,1,4)(7,2,3)(7,3,2)(7,4,1) - 4
Hence total no.of possible solutions - 37 - 11 years agoHelpfull: Yes(22) No(5)
- possible combinations and their different arrangements are:
(1,4,7)=3!
(1,5,6)=3!
(2,3,7)=3!
(2,4,6)=3!
(2,5,5)=3
(3,4,5)=3!
(3,6,3)=3
(4,4,4)=1
no. of possible solutions= 3!+3!+3!+3!+3+3!+3+1
= 6+6+6+6+3+6+3+1
= 37 - 10 years agoHelpfull: Yes(15) No(0)
- Fix a at 1:
possible set of (n,p)=(1,10),(2,9)....(10,1) =10
at a=2
possible set of (n,p)=(1,9),(2,8)....(9,1) =9
.
.
.
at a=7
possible set of (n,p)=(1,4),(2,3)....(4,1) =4
So ans=10+9+8+7+6+5+4
=49(ans) - 11 years agoHelpfull: Yes(6) No(28)
- answer=55
- 11 years agoHelpfull: Yes(3) No(3)
- a,n,p be have values 1,4,7 -> 3ways
1,5,6 ->3 ways
2,3,7->3ways
2,4,6 ->3
2,5,5 ->3
3,3,6 ->3
3,4,5 ->3
4,4,4 ->1
so,total of 22 ways - 11 years agoHelpfull: Yes(3) No(4)
- 18 ways
1. 1,4,7 - 3! ways
2. 1,5,6 - 3! ways
3. 1,6,5 - 3! ways
hence 3!+3!+3! = 18 ways - 11 years agoHelpfull: Yes(2) No(11)
- If repeatition is allowed then 37 and if not then 30
- 10 years agoHelpfull: Yes(2) No(0)
- repetition of no. is allowed?
- 11 years agoHelpfull: Yes(1) No(1)
- Gaurav Nagdev's solution is nice but requires more time :p Sorry, dude !!!
- 11 years agoHelpfull: Yes(1) No(1)
- 9 solutions
2+3+5(3tyms)
2+4+6(3 tyms)
3+4+5(3 tyms) - 11 years agoHelpfull: Yes(0) No(12)
- 30 solutions.....
- 11 years agoHelpfull: Yes(0) No(3)
- 5 solutions
7+4+1
7+3+2
6+5+1
6+4+2
5+4+3
- 11 years agoHelpfull: Yes(0) No(3)
- 0,4,8-3! ways
0,5,7-3!
1,4,7
2,3,7
2,4,6
3,2,7
3,4,5
6,5,1
6,4,2
so 9*6= 54 ways.. - 11 years agoHelpfull: Yes(0) No(2)
- 38 is
ans
- 11 years agoHelpfull: Yes(0) No(0)
- (1+4+7),(1+5+6),(1+6+5),(1+7+4)
(2+3+7),(2+4+6),(2+5+5),(2+6+4),(2+7+3)
so on...
total no of solution is 49 - 10 years agoHelpfull: Yes(0) No(0)
- If we consider only positive solution then
x1+x2+x3+....+xk=n
then no of solution
(n-1)
C
(k-1)
if zero is included then non-negative positive solution
(n+k-1)
C
(k-1)
For this case n=12 and k=3
then only positive solution
(12-1)
C
(3-1)
which gives
55. - 10 years agoHelpfull: Yes(0) No(0)
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