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Numerical Ability
Permutation and Combination
how many ways does 10 pencils given to four children such that each get atleast 1?
Read Solution (Total 24)
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- Ans is:84
(10-1)C(4-1)=9C3=9*8*7/3*2=504/6=84 - 10 years agoHelpfull: Yes(44) No(4)
- There are 10 pencils
so 1 given to each
hence 6 remains
as per product rule and sum rule the total ways will be
4*(6+5+4+3+2+1)=84 - 10 years agoHelpfull: Yes(14) No(1)
- Formula is (n-1)C(r-1)
- 10 years agoHelpfull: Yes(12) No(3)
- Number of ways of distributing n identical things among r person where Each person get atleast one...
- 10 years agoHelpfull: Yes(8) No(1)
- in which case we use this formula..? only when we have to distribute atleast one to all ?
- 10 years agoHelpfull: Yes(1) No(1)
- @karthikeyank plz explain
- 10 years agoHelpfull: Yes(1) No(0)
- (10-1)C(4-1)
- 10 years agoHelpfull: Yes(1) No(1)
- pencils are same and 1 children get 1 pencil
remaining pencils are 6
so
1c=6pcncils
2c=6pencils
3c=6pencils
4c=6pencils
answer is 6^4=1296 - 10 years agoHelpfull: Yes(1) No(8)
- 4^10
1 pencil can be given to 4 children in 4 ways.
similary 10 pencils in 4^10 - 10 years agoHelpfull: Yes(1) No(3)
- (10-1)C(4-1)=84
- 10 years agoHelpfull: Yes(1) No(0)
- @karthikey- plz explain the solution, and how to use this formula
- 10 years agoHelpfull: Yes(0) No(0)
- @Karthik
plz explain the answer - 10 years agoHelpfull: Yes(0) No(1)
- i think ans is 10C4
- 10 years agoHelpfull: Yes(0) No(7)
- note:if pencil are similar.
first give 1 pencil to each children.
now as 6 pencil are remaining.
now 1st pencil out of 6 can be given to any 4 children i.e. there are four ways to give 1 pencil to 4 children.
-->similarly, 2nd pencil can be given to any of 4 children.. and so on..upto 6th pencil.
thus we get 4^6 ways .
which is 4^6=256.
- 10 years agoHelpfull: Yes(0) No(5)
- (n-1)C(r-1)
Its a concept we use when u have to distribute n similar item between r persons such that all get at least 1. So. 9C3 - 10 years agoHelpfull: Yes(0) No(0)
- can any one say without using formulae?????
- 10 years agoHelpfull: Yes(0) No(0)
- number of ways distributing n identical things among p groups so that atleast each get one = (n-1)c(p-1)
so ways= 9c3 - 10 years agoHelpfull: Yes(0) No(0)
- (10-1)C(4-1)=9C3=84
- 10 years agoHelpfull: Yes(0) No(0)
- distribution of n things in n numbers hat every one get atleast one item =
(n-1)C(r-1)
so we get (10-1)C(4-1)=84 - 10 years agoHelpfull: Yes(0) No(0)
- if we give 1 pencil to four children ... after that 6 pencils are remaining .... thus .. we can ditribute .. these in ..= 6*6*6*6=256 ways..
- 10 years agoHelpfull: Yes(0) No(0)
- 10c4-4c2/10c4
- 10 years agoHelpfull: Yes(0) No(0)
- plz explain clearly
- 10 years agoHelpfull: Yes(0) No(0)
- (10-1)C(4-1)=84
- 10 years agoHelpfull: Yes(0) No(0)
- (n-1)C =9C3
(r-1) - 10 years agoHelpfull: Yes(0) No(0)
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