Latest TCS Aptitude Question SOLUTION: Seven different objects must be decided among 3 persons. In how many ways this can be done if at least one of them gets exactly one object???

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Seven different objects must be decided among 3 persons. In how many ways this can be done if at least one of them gets exactly one object???

Asked In TCS A M ANKIT KALLURAYA (9 years ago)

Unsolved Read Solution (21)

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Read Solution (21) :

rahul  (9 years ago)

n-1c r-1 =6c2=15

Is this solution Helpfull? Yes (15) | No (7)  

Emanuel Abhishek Murmu  (9 years ago)

Division of m+n+p objects into three groups is given by (m+n+p)/(m!×n!×p!)
so, according to question at least one gets only one object this can be done in (1,1,5),(1,2,4),(1,3,3) ways.

required number of ways= (7!/(1!*1!*5!))*(1/2!) + (7!/(1!*3!*3!))*(1/2!) + (7!/(1!*2!*4!)) = 196

Is this solution Helpfull? Yes (11) | No (3)  

Surit Datta  (9 years ago)

(1) suppose only 1 person gets 1 object.
so, the other two get either (0,6) or (2,4) or (3,3) objects.
so, no. of ways of partitioning the remaining 6 objects into each such group = 1, 15, 20 respectively
req no. of ways this can happen = 7*3*(1+15+20)*2

(2) now suppose only 2 persons get 1 object each
so, the third must get remaining 5
req no. of ways this is possible = 3*21

add up the nos.

Is this solution Helpfull? Yes (2) | No (2)  

JIT BHATTACHARJEE  (8 years ago)

(7!/(1!*1!*5!))*(1/2!) + (7!/(1!*3!*3!))*(1/2!) + (7!/(1!*2!*4!)) = 196

Is this solution Helpfull? Yes (2) | No   

Deep  (9 years ago)

Answer must be like in that way...
They have already said different object so first take a,b,c three people ,a get 1 object in 7 way .and remaining 6 obj can be divided as 6c2 way .as they diffrnt ..so thats 7x6c2=105 ways..and this can be happened three way.. so ans in 315 .

Is this solution Helpfull? Yes (1) | No (2)  

santhiya  (9 years ago)

Division of m+n+p objects into three groups is given by (m+n+p)!m!*n!*p!
But 7=1+3+3 or 1+2+4 or 1+1+5
so the number of ways are (7!) 1! * 3!*3!*12!+(7)!1!2!4!+(7)!1!*1!*5!*12!=70+105+21=196
so the answer is 196

Is this solution Helpfull? Yes (1) | No (1)  

rahul kataria  (9 years ago)

formula for this is( m+n+p)!/m!n!p!
so 7!/3!=140 ways

Is this solution Helpfull? Yes | No (4)  

Arka Paul  (9 years ago)

7 different objects must be decided among 3 persons.
so,no of ways to get by exactly one person,exactly one object = 7C3 -1

Is this solution Helpfull? Yes | No (1)  

md abdulla choush  (9 years ago)

7c3=7*6*5*4*3/1*2*3
=420

Is this solution Helpfull? Yes | No (1)  

SUMAN GHOSH  (9 years ago)

6 items can be divided into 6c2=15ways
and anyone gets one item
so total no ways =15*3=45

Is this solution Helpfull? Yes | No (1)  

vartika  (9 years ago)

7p3 is the right answer

Is this solution Helpfull? Yes | No (2)  

kashik garg  (9 years ago)

7c3 - 3 - 0 = 7c3 - 3
7c3 = 7 different objects decided among 3 persons.
3 = ways in which 1 object can be given to 2 men.
0 = ways in which 1 object can be given to 3 men
So the remaining are the ways in only 1 man get 1 object

Is this solution Helpfull? Yes | No   

Eswar Mohan  (9 years ago)

(1) suppose only 1 person gets 1 object.
so, the other two get either (0,6) or (2,4) or (3,3) objects.
so, no. of ways of partitioning the remaining 6 objects into each such group = 1, 15, 20 respectively
req no. of ways this can happen = 7*3*(1+15+20)*2

(2) now suppose only 2 persons get 1 object each
so, the third must get remaining 5
req no. of ways this is possible = 3*21

Is this solution Helpfull? Yes | No (1)  

Mohamed Mukthar  (9 years ago)

lets take it like this _ _ _ .3 people have to get from 7 objects. In question it has stated that one has to get exactly one object so 1st person gets one then remaining 2 person has to get from 6 objects

possibilities
2nd person 6
3rd person 5
so 1*60*5 = 30

Ans:30

Is this solution Helpfull? Yes | No   

Nikhil Bharatkar  (9 years ago)

The objects can be divided into three groups that are (1,1,5) or (2,4,1) or (3,3,1) as atlas one person should get 1 object.
so number of ways can be [(7C1*6C1*5C5)/2!]+(7C2*5C4*1C1)+[(7C3*4C3*1C1)/2!]=196

Is this solution Helpfull? Yes | No   

shaik salman  (8 years ago)

case1: * * *
1 object for one person and remaining 6 objects for 2 persons=6c2=15
three cases are possible so 3*15=45
case2: for 2 persons each one object then fr one person 5 objects
3 cases are possible so 3*5=15
hence 15+45=60

Is this solution Helpfull? Yes | No   

kavana j  (8 years ago)

there 7 objects
so 1 give to each
4 remains
as per product and sum rule 3*(4+3+2+1)=3*10=30
ans:30

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Tathagata Mondal  (8 years ago)

there are 3 possible dist. scenario. (1,1,5),(1,2,4),(1,3,3)
(1,1,5)
since seven objects are different, not identical, thus no of ways of dist.ing them = (7C1+6C1+5C5)*(3!/2!)=126
similarly,
for(1,3,3)
(7C1*6C3*3C3)*(3!/2!)=420
for(1,2,4)
(7C1*6C2*4C4)*(3!)=630
hence total ways=(630+420+126)=1176

Is this solution Helpfull? Yes | No   

VIVEK JAIN  (8 years ago)

Answer is 15
because
x+y+z=7
question says at least one
(1,1,5)=this can be done in 3!/2!=3 or(+)
(2,2,3)=this can be done in 3!/2!=3 or(+)
(1,3,3)=this can be done in 3!/2!=3 or(+)
(1,2,4)=this can be done in 3! = 6
so finally sum is 15
:)

Is this solution Helpfull? Yes | No   

vinav kumar agrawal  (8 years ago)

1 2 4 3! =6
1 3 3 3!/2=3
1 1 5 3!/2=3
6+3+3=12

Is this solution Helpfull? Yes | No   

MANISH DHYANI  (7 years ago)

ans is 18 ..because atliest one of them get one object!!!

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