Elitmus
Exam
Numerical Ability
Permutation and Combination
Compute the no. of distinct ways in which 56 toffees can be distributed to A,B,C,D,E so that no person receives less than 10 toffees?
A)10c6 B)10c5 C)6^3
Read Solution (Total 12)
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- Distribute 10 toffees to each of 5 persons. so total 50 toffees , now remaining toffees are 6 which we hav to distribute among 5 person so this can be possible in
6+5-1C5-1=10C4=10C6 so option will be A
- 11 years agoHelpfull: Yes(68) No(5)
- distribute 10 toffes to each first: after that 6 toffes (r),5 persons(n)
(n+r-1)C(r-1)=10C5 - 11 years agoHelpfull: Yes(25) No(32)
- NO PERSON RECEIVES LESS THAN 10 TOFFEES
so 50 toffees are distributed equally i.e 10 toffees to each of A,B,C,D,E
now (56-50=6) toffees are to be distributed among 5 persons
we know that
number of ways in which n identical things can be distributed among r persons, each one of them can get 0,1,2 or more
=(n+r-1)C(r-1)
here n=6(toffee) is to distributed among r=5(persons)
no. of ways=(6+5-1)C(5-1)=10C4=10C6 - 10 years agoHelpfull: Yes(17) No(0)
- @akanksha...
this question was asked in elitmus...?? - 11 years agoHelpfull: Yes(15) No(8)
- (a+10)+(b+10)+(c+10)+(d+10)+(e+10)=56(all of them will receive atleast 10 toffee)
a+b+c+d+e=6 (we distribyted 6 among 5 )total no. of ways 10c4=10c6 - 9 years agoHelpfull: Yes(2) No(0)
- distribute 10 coffee to each so total 50 toffees,and now remaining toffee are 6 and now we have to distribute 6 toffee in 5 person.let n=6,r=5 so total way is (n+r-1)C(r-1)=(6+5-1)C(5-1)
=10C4.
why i used this formula you can understand from this explanation.
Consider that the number of identical objects as n.Now consider that all the n objects are placed on a row.Let us assume some separators that separates the n objects.Now as we want to separate them into r group[as blank groups are also allowed],take the separators as objects.Now we need r-1 separators to make r groups.Therefore total number of objects is n+r-1.There will be r-1 places for the separators to occupy.Therefore we can arrange the separators in
(n+r-1)C(r-1) ways, which is the answer to the question asked. - 8 years agoHelpfull: Yes(2) No(0)
- first we will distribute 10 toffees to each of 5 persons i.e total 50 toffees ,
now remaining toffees are 6 which we have to distribute among 5 person so this can be possible in
m+r-1 C r-1
=>6+5-1C5-1=10C4 - 10 years agoHelpfull: Yes(1) No(0)
- option 4 is 5^6
the ans is 5^6 - 7 years agoHelpfull: Yes(1) No(2)
- samjh nahi aaya...bhai solution...
- 10 years agoHelpfull: Yes(0) No(0)
- option a.is the ans
- 9 years agoHelpfull: Yes(0) No(0)
- NO PERSON RECEIVES LESS THAN 10 TOFFEES
so 50 toffees are distributed equally i.e 10 toffees to each of A,B,C,D,E
now (56-50=6) toffees are to be distributed among 5 persons
we know that
number of ways in which n identical things can be distributed among r persons, each one of them can get 0,1,2 or more
=(n+r-1)C(r-1)
here n=6(toffee) is to distributed among r=5(persons)
no. of ways=(6+5-1)C(5-1)=10C4=10C6
Read more at http://www.m4maths.com/17628-Compute-the-no-of-distinct-ways-in-which-56-toffees-can-be-distributed-to-A-B-C-D-E-so.html#3GwjU8Y8ft7ZtyiF.99 - 9 years agoHelpfull: Yes(0) No(4)
- a,b,c,d,e>10
let us assume k>=1
now a,b,c,d,e,k>=11
So 11-1C6-1=10C5 - 7 years agoHelpfull: Yes(0) No(3)
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