Latest TCS Aptitude Question SOLUTION: IN how many ways we can distribute 10 identical looking pencils to 4 students so that each student get at least one pencil? a)5040 b)210 c)84 d)none of these

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(#M40010644) TCS Numerical Ability question Keep an EYE

IN how many ways we can distribute 10 identical looking pencils to 4 students so that each student get at least one pencil?
a)5040
b)210
c)84
d)none of these

Asked In TCS (11 years ago)

Unsolved Read Solution (14)

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Read Solution (14) :

ragini devi  (11 years ago)

consider 4 students as A,B,C,D THEN A+B+C+D=10, but each one should get atleast 1 pencil so giving 1 pencil to each student, the remaining pencils are 10-4=6
now A'+B'+C'+D'=6, number of non negative integers of A',B',C',D' ARE (6+4-1)C(4-1) I.E COMBINATIONS, 9c3=84, c is correct answer

Is this solution Helpfull? Yes (78) | No (12)  

ragini devi  (11 years ago)

we have formulas regarding such sort of distributions:
x1+x2+x3......xn=r, then 1)the number of +ve integral solutions(excluding zero)is (n-1)c(r-1) and 2)the number of -ve integral solutions (including zero) is (n+r-1)c(r-1)

Is this solution Helpfull? Yes (24) | No (4)  

Anshul Gupta  (11 years ago)

a1+a2+a3+....+ar=n
1) No. of non negative integral solutions = n+r-1 C r-1
2) No. of positive integral solutions : n-1 C r-1
so in given question
we have r=4(students having pencils)
sum of all the pencils of students is 10
so
a1+a2+a3+a4=10
No. of positive integral solutions : n-1 C r-1
(as each student have atleast 1 pencil)
10-1 C 4-1= 36

Is this solution Helpfull? Yes (10) | No (26)  

sudhakar jain  (10 years ago)

firstly give 1 pencil each to the 4, now we can distribute the remaining 6 pencils any way we like.
using the "stars and bars" formula, (6+4-1)C(4-1)
= 9C3 = 84 [ c ]

Is this solution Helpfull? Yes (9) | No   

venkatesh  (11 years ago)

(n+r-1)c(r-1)

Is this solution Helpfull? Yes (7) | No (4)  

Lovejeet Singh  (11 years ago)

@Ragini: Plzz explain how this come:
A'+B'+C'+D'=6, number of non negative integers of A',B',C',D' ARE (6+4-1)C(4-1) I.E COMBINATIONS, 9c3=84
How u GOT (6+4-1)C(4-1) ???
Thx in Advance !!!! :) :)

Is this solution Helpfull? Yes (6) | No (4)  

disha verma  (11 years ago)

ragini ji,plz elaborate it....

Is this solution Helpfull? Yes (5) | No (3)  

Naveenkumar Selvaraj  (11 years ago)

10C4=210
Out of 10 pencils 4 are given to student each

Is this solution Helpfull? Yes (4) | No (67)  

Arpit goel  (9 years ago)

answer=84
Number of ways in which n identical things can be distributed among r persons, each one of whom can receive 1,2 or more items (n-1)C(r-1).
so answer =9c3=84,c is correct

Is this solution Helpfull? Yes (4) | No (2)  

Rajanikanta Pradhan  (8 years ago)

ans 84
n-1 c r-1 as at least one so non of them get zero

Is this solution Helpfull? Yes (3) | No   

Mohd. Shahnawaz Husain  (8 years ago)

The no. of ways in which 'n' identical things can be divided in 'r' groups when each person get any no. of thing including zero= (n+r-1)C(r-1)
And
The no. of ways in which 'n' identical things can be divided in 'r' groups when each person get atleast one thing = (n-1)C(r-1)



(10-1) C (4-1)= 9! / ( 6! * 3! ) = 84. c is the correct answer.

Is this solution Helpfull? Yes (3) | No   

raj kumar  (11 years ago)

Ans:b)210
Solution:
=(10*9*8*7)/(4*3*2*1)
=210

Is this solution Helpfull? Yes (2) | No (49)  

Prateek Nanhorya  (10 years ago)

qdssdfdsfadsf
dsffsadfsf

Is this solution Helpfull? Yes (2) | No (9)  

ammu  (8 years ago)

the formula is (n-1)C(r-1)= 9C3= 84

Is this solution Helpfull? Yes (2) | No   

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