Latest TCS Aptitude Question SOLUTION: 10 nos of different gifts have to distributed among 4 people so that every one get at-least 1 gifts..

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10 nos of different gifts have to distributed among 4 people so that every one get at-least 1 gifts..

Asked In TCS agnimitra das (9 years ago)

Unsolved Read Solution (17)

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Read Solution (17) :

Palak Gupta  (9 years ago)

(10-1)C(4-1)= 84

Is this solution Helpfull? Yes (36) | No (4)  

BHASKAR MANDAL  (9 years ago)

as the total is 10 no. of gifts different possibilities are
(7,1,1,1) = 4!/3! = 4
(6,2,1,1) = 4!/2! =12
(5,3,1,1) = 4!/2! =12
(5,2,2,1) = 4!/2! =12
(4,4,1,1) = 4!/(2!*2!) = 6
(4,3,2,1) = 4! =24
(4,2,2,2) = 4!/3! = 4
(3,3,3,1) = 4!/3! = 4
(3,3,2,2) = 4!/(2!*2!) = 6

total possibilities is the addition of the above = 84

Is this solution Helpfull? Yes (23) | No (5)  

shilpi singh  (9 years ago)

as gifts r distinct so each price can b
given in 4 ways so ans will b 4^10 and excluding d condition of getting 0 gifts. ans is 4^10-1

Is this solution Helpfull? Yes (15) | No (1)  

rahul garg  (9 years ago)

(10+4-1)C(4-1)=13C3

Is this solution Helpfull? Yes (12) | No (4)  

Aastha  (9 years ago)

for distributing n identical thing among r people,so that each get atleast 1 is
n-1Cr-1
9C3=42

Is this solution Helpfull? Yes (6) | No (17)  

lokesh sharma  (9 years ago)

in such hind of question use the formula n-1cr-1

Is this solution Helpfull? Yes (5) | No   

Rohit Shaw  (9 years ago)

the gifts are different so we can arrange in 4^10 ways ,
but due to the restriction that every one get at-least one.
select 4 from 10 diff. gifts in 10C4 way and we can arrange this in 4! way.
The remaining gifts are distributed in 4^6 ways without restriction.
hence the naser will be : 10C4*4!*4^6.

Is this solution Helpfull? Yes (4) | No (1)  

fathima mehajabin  (9 years ago)

10C4 is the correct answer.9C3 will be correct if and only if it was identical gifts.

Is this solution Helpfull? Yes (4) | No (1)  

Rahul kumar choudhary  (9 years ago)

using partition rule,
when each people has to have atleast one thing.
(n-1)C(r-1)
therefore,
9C3=84

Is this solution Helpfull? Yes (4) | No (1)  

Archana Gehlot  (9 years ago)

(n-1)c(r-1) 9c3 =84ans

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

Archana Gehlot  (9 years ago)

(n-1)c(r-1) 9c3 =84

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

neha kumari  (9 years ago)

since everyone sud get atleast 1 gift
so
s+x+y+z=10-(4)
s+x+y+z=6
(6+4-1)c(4-1)=84

Is this solution Helpfull? Yes (1) | No   

Pratik sharma  (9 years ago)

formula is (n-k)c(r-k)...n->total object
r->here stand for no of people
k-> k no of object always get choosen

so (10-1)c(4-1)=9C3=84

Is this solution Helpfull? Yes (1) | No   

sarang s naik  (9 years ago)

13c4=13!/9!4!

Is this solution Helpfull? Yes | No   

Pragna Paramita Pal  (9 years ago)

Since the question is to distribute different gifts so 4 gifts can be selected from 10 gifts in 10C4 ways such that each has atleast 1 gift and this can be arranged in 4 ways Now 6 gifts are left to be distributed and this can be distributed inwhen only 1 gets all the 6 gifts- 4 wayswhen two gets- 6c51c16c42c26c33c362when three of them get- 6c42c11c16c33c21c16c24c22c264when all 4 of them get-6c33c12c11c16c24c22c11c124now summing up all the types of distribution 10c4446c51c16c42c26c33c3626c42c11c16c33c21c16c24c22c2646c33c12c11c16c24c22c11c124

Is this solution Helpfull? Yes | No (1)  

kundan karn   (9 years ago)

ANS IS 1
55/3=1
55/3^2=1
55/3^3=1

Is this solution Helpfull? Yes | No (1)  

Rahul Das  (9 years ago)

4*(6+5+4+3+2+1)=84

Is this solution Helpfull? Yes | No   

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