Latest CAT Aptitude Question SOLUTION: 7 different objects must be divided among 3 people.In how many ways can this be done if atleast one of them gets exactly one object

It is mentioned that objects are DIFFERENT and also atleast one of them also has to get exactly 1 object.
Now, since it is not explicitly mentioned in the problem we assume that no1 gets zero object.
We can choose 3 objects from 7 in (7C3 ways)= 35 ways.
Now, for each of these 35 cases , the number of ways in which these chosen 3 objects can be distributed among 3 persons so that every1 gets only 1 object = 3! =6 ways.
So, total number of ways in which 3 from 7 different objects can be distributed among 3 persons so that every person gets only 1 object = 35*6 = 210 ways ..

Now,for each of these 210 cases,2 persons from 3 can be chosen in 3C2 =3 ways..
So, totally we get (210*3)cases=630 cases
For each of these 630 cases ,we can distribute the remaining 4 objects to these 2 chosen persons so that each of them gets 2 of objects in
= (4C2)= 6 ways.

So, the number of ways in which 7 diff. objects can be distributed among 3 persons so that ONLY ONE (mind it !!) of them gets exactly 1 object in = 630*6
= 3780 ways.............[result 1]

Here, we have to consider another case when 2 persons get 1 object each.
firstly we know,
We can choose 3 objects from 7 in (7C3 ways)= 35 ways.
Now, for each of these 35 cases , the number of ways in which these chosen 3 objects can be distributed among 3 persons so that every1 gets only 1 object = 3! =6 ways.
So, total number of ways in which 3 from 7 different objects can be distributed among 3 persons so that every person gets only 1 object = 35*6 = 210 ways .
Now, for each of these 210 cases ,we can choose a person from the 3 so that the chosen person is given the remaining 4 objects, in = 3C1 = 3 ways

So, total number of ways in which 2 persons get 1 object each and the other gets the remaining 5 , = 210*3 = 630.............[result 2]

Now, since [result 1] & [result 2] are generated from two mutually exclusive cases,i.e, since these two cases have no over-lapping region, so total number of ways in which 7 different objects must be divided among 3 people so that atleast one of them gets exactly one object are = [result 1] + [result 2] = 3780 + 630 = 4410 ways..

So, the given task with the given condition can be done in 4410 ways..

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Consider all division combinations
A(1) B(1) C(5) = 3!*7!/(2!*5!) = 126
3! since A,B,C can be interchanged among themselves, 2! to account for same no. of objects in group A,B
Similarly,
A(1) B(2) C(4) = 3!*7!/(2!*4!) = 630
A(1) B(3) C(3) = 3!*7!/(2!*3!*3!) = 420
For total ways, Add all to get ans as 1176

I think Ans. will be 1218 because here one more case is also possible(AS PER QUESTION) i.e.,
A can take 1 object
B can take 0 object
C can take 6 objects
so ways will be 3!*(1+0+6)!/(6!)= 6*7=42
AND 1176 will come, Consider all division combinations
A(1) B(1) C(5) = 3!*7!/(2!*5!) = 126
3! since A,B,C can be interchanged among themselves, 2! to account for same no. of objects in group A,B
Similarly,
A(1) B(2) C(4) = 3!*7!/(2!*4!) = 630
A(1) B(3) C(3) = 3!*7!/(2!*3!*3!) = 420
For total ways, Add all to get ans as (42+126+630+420)=1218 :)

Firstly we distribute 3 object each get exactly one.
so remain is 4.
so according to partision formula n+r-1cr-1
(4+3-1)C(3-1)=6C2=15 ways

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

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

Firstly we distribute 3 object each get exactly one.
so remain is 4.
so according to partision formula n+r-1cr-1
(4+3-1)C(3-1)=6C2=15 ways

if only 1 gets 1 object , the remaining can be distributed as (6,0),(4,2), (3,3).
(7C1 X 6C6 X 3! + 7C1 X 6C3 X 3!2!)= 42+630+420+1092.

IF 2 PEOPLE GET 1 OBJECT EACH
(7C1 X 6C1 X 5C5 X 3!2!)=126

THUS A total of 1218

196 is d correct one al other are wrng

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!)×(1/2!)+(7!/(1!×2!×4!)+(7!/(1!×1!×5!)×(1/2!) = 70 + 105 + 21 = 196

Division of m+n+p objects into three groups is given by (m+n+p)!m!×n!×p!(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!(7)!1!×3!×3!×12!+(7)!1!×2!×4!+(7)!1!×1!×5!×12! = 70 + 105 + 21 = 196

Firstly we distribute 3 object each get exactly one.
so remain is 4.
so according to partision formula n+r-1cr-1
(4+3-1)C(3-1)=6C2=15 ways

Firstly we distribute 3 object each get exactly one.
so remain is 4.
so according to partision formula n+r-1cr-1
(4+3-1)C(3-1)=6C2=15 ways

but from where does that 12! come?

It is mentioned that objects are DIFFERENT and also atleast one of them also has to get exactly 1 object. so for selecting 1 object from 7 different things and giving to one of 3 persons can be done in 7C1*3C1 ways. there are remaining 6 different objects. each can be given two either of two guys so it is 2*2*2*2*2*2=2^6 ways( includes the situation where zero objects can be given to one).

So there are 7C1*3C1*(2^6) = 1344 ways.

it can divide in (1,3,3),(1,2,4),(1,1,5)
so 1/2*7!/(3!*3!)+1/2*7!/(2!*4!)+1/2*7!/5!
=196

Firstly we distribute 3 object each get exactly one.
so remain is 4.
so according to partision formula n+r-1cr-1
(4+3-1)C(3-1)=6C2=15 ways

Firstly we distribute 3 object each get exactly one.
so remain is 4.
so according to partision formula n+r-1cr-1
(4+3-1)C(3-1)=6C2=15 ways

Firstly we distribute 3 object each get exactly one.
so remain is 4.
so according to partision formula n+r-1cr-1
(4+3-1)C(3-1)=6C2=15 ways

Firstly we distribute 3 object each get exactly one.
so remain is 4.
so according to partision formula n+r-1cr-1
(4+3-1)C(3-1)=6C2=15 ways

Firstly we distribute 3 object each get exactly one.
so remain is 4.
so according to partision formula n+r-1cr-1
(4+3-1)C(3-1)=6C2=15 ways

ans 1176
let
a=1 obj=7
b=1 obj=15
c=5 obj=1 now7*15=105 and 105*3=315

similarly

can anyone explain me what is the correct answer all the solutions which you guys are posting is increasing my confusion

196 is correct

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

This question has same concept like the question(In how many ways can 6 persons be accommodated in 4 different rooms such that no room remains empty).To understand better, follow this link(http://2iim.com/Combinatorics%20from%20QA4CAT.pdf, que no is 17)).So the correct answer is 1176.

1218 laga raho

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! ×1/2! + (7)!/1!×2!×4! + 1/2! * (7)! / 1!×1!×5! = 70 + 105 + 21 = 196

here one more case also possible that is 2+2+3. why it has been left in each answer given by you all.please help