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Numerical Ability
Permutation and Combination
what are the number of solutions for x+y+z=12, if x,y & z can take values from 1 to 7 ?
Read Solution (Total 12)
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- Ans=37
if x=1, (y,z) can be (4,7)(5,6)(6,5)(7,4)
if x=2, (y,z) can be (3,7)(4,6)(5,5)(6,4)(7,3)
if x=3, (y,z) can be (2,7)(3,6)(4,5)(5,4)(6,3)(7,2)
if x=4, (y,z) can be (1,7)(2,6)(3,5)(4,4)(5,3)(6,2)(7,1)
if x=5, (y,z) can be (1,6)(2,5)(3,4)(4,3)(5,2)(6,1)
if x=6, (y,z) can be (1,5)(2,4)(3,3)(4,2)(5,1)
if x=7, (y,z) can be (1,4)(2,3)(3,2)(4,1)
Total Combination = 4+5+6+7+6+5+4=37 - 11 years agoHelpfull: Yes(68) No(2)
- Sorry I was wrong.Its 37 ways..
Possiblities are
1 4 7
1 5 6
2 3 7
2 4 6
3 4 5
2 5 5
3 3 6
4 4 4
These 7 cases will add up to 12.
Btw take the first case 1 4 7
x=1 y=4 z=7
x=1 y=7 z=4
x=4 y=1 z=7
x=4 y=7 z=1
x=7 y=4 z=1
x=7 y=1 z=4
Thus for the first case, 6 permutations are possible..
for the first 5 cases , 6*5 =30 permutations are possible.
But for 3 3 6 and 2 5 5 repetition is there. (3!/2!)*2 = 6
And for 4 4 4, 1 permutation.
So
30+6+1=37 - 11 years agoHelpfull: Yes(10) No(3)
- The correct answer is 37.
- 11 years agoHelpfull: Yes(4) No(1)
- I think 42 ways.. I couldn't solve in any easy method.
Possiblities are
1 4 7
1 5 6
2 3 7
2 4 6
2 5 5
3 3 6
3 4 5
These 7 cases will add up to 12.
Btw take the first case 1 4 7
x=1 y=4 z=7
x=1 y=7 z=4
x=4 y=1 z=7
x=4 y=7 z=1
x=7 y=4 z=1
x=7 y=1 z=4
Thus for the first case, 6 permutations are possible..
we have 7 cases,so 6*7=42 - 11 years agoHelpfull: Yes(3) No(15)
- I made a mistake...37 is correct answer
- 11 years agoHelpfull: Yes(1) No(2)
- we can write as 7+4+1/7+3+2 smilarly we can write n get 9 solutn
- 11 years agoHelpfull: Yes(0) No(7)
- 12+3-1 C 12=91.
- 11 years agoHelpfull: Yes(0) No(10)
- 43 solutions are possible
- 11 years agoHelpfull: Yes(0) No(12)
- (1,4,7)---->can be arranged in 6 ways to give a solution
(1,5,6)---->can be arranged in 6 ways to give a solution
(2,3,7)---->can be arranged in 6 ways to give a solution
(2,4,6)---->can be arranged in 6 ways to give a solution
(2,5,5)---->can be arranged in 3 ways to give a solution
(3,3,6)---->can be arranged in 3 ways to give a solution
(3,4,5)---->can be arranged in 6 ways to give a solution
(4,4,4)---->can be arranged in 1 way to give a solution
(5,5,2)---->can be arranged in 3 ways to give a solution
Total no of ways= 40 - 11 years agoHelpfull: Yes(0) No(14)
- sorry for previous answer it will be 37
- 11 years agoHelpfull: Yes(0) No(3)
- @KARTHIKA why didnt u take 5 5 2???
- 11 years agoHelpfull: Yes(0) No(1)
- @Rudra:
x=2 y=5 z=5
x=5 y=5 z=2
x=5 y=2 z=5
these are the three ways...
So I just tuk 2 5 5 and did permutation. i.e 3!/2!=3ways as mentioned in detail above (Dividing by 2!,because "5" is repeated two times)
Rudra hope yu understud
- 11 years agoHelpfull: Yes(0) No(0)
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