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Numerical Ability
Permutation and Combination
in how many ways 10 identical looking pencils to 4 students so that each student get atleast one pencil
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- there are 10 pencils so 1 give to each
hence 6 pencil remains
4*(6+5+4+3+2+1)
=84 - 11 years agoHelpfull: Yes(15) No(1)
- 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 - 11 years agoHelpfull: Yes(14) No(2)
- First distribute 1 pencil each to 4 students. So remaining pencils are 6.
Then ways of distributions are (n+r-1)C(r-1)
Here n=6 & r=4
hence 9C3=84 - 11 years agoHelpfull: Yes(3) No(0)
- 6C4=15
after giving 1 pencil to each we have 6 pencils left which can be distributed in 6C4 ways. - 11 years agoHelpfull: Yes(3) No(1)
- formula is n-1Cr-1,,,, 9C3
- 9 years agoHelpfull: Yes(1) No(0)
- ans=84
logic is==(n-1)C(r-1)=9C3=9*8*7/3*2*1=84 - 11 years agoHelpfull: Yes(0) No(0)
- by natural number solution method, n=10,r=4, so we have have n-1Cr-1 i.e. 9C3 is the answer
- 9 years agoHelpfull: Yes(0) No(0)
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