Latest TCS Aptitude Question SOLUTION: Zada has to distribute 15 choclates among 5 of her children sana,ada,amir,farhan. she has to make sure that sana get atleast 3 and atmost 6 choclates.in how many ways can this be d

there would be 4 cases
1.A gets 3 chocolates then the other 12 can be distributed among other 4 children in (12-1)C(4-1)ways ,ie;165 ways
2.A gets 4 chocolates then the other 11 can be distributed among other 4 children in (11-1)C(4-1)ways ,ie;120 ways
3.A gets 5 chocolates then the other 10 can be distributed among other 4 children in (10-1)C(4-1)ways ,ie;84 ways
4.A gets 6 chocolates then the other 9 can be distributed among other 4 children in (9-1)C(4-1)ways ,ie;56 ways

adding all above cases,we get 165+120+84+56=425 ways.....

hence the solution is 425 ways.

In this we don't use the formula (n-1)c(r-1). Because this formula is distributing
n identical things among r people/groups such that any person/group might get any number(except 0).
But here the others children might get zeros no. of chocolate.
So we use the formua (n+r-1)c(r-1). so we get the answer 888.

435 ways of distributing the chocolates

you people are assuming that each child gets atleast 1 chocolate but it can be a case that a child get no chocolate..so in my opinion the answer would be 15c3+14c3+13c3+12c3
what you guys say???

sana gets 3,and then we give 1 each to remaining 4 so, 7 chocolates over we have 8 chocolates remaining.now we need to distribute the remaining 8 to the other four children.this can be done in 8+4-1C4-1 ways [case 1]. In line with this we also have cases 2,3,4.

case 1 when sana gets 3 chocolates: ( 8+4-1)C(4-1) = 11C3 = 165

case 2 when sana gets 4 chocolates 10C3 = 120

case3 sana gets 5 chocolates 9C3 = 84

case4 sana get 6 chocolates 8C3 = 56

thus total ways are 165 + 120+ 84+56 = 425

11c3+10c3+9c3+8c3=425

is it actually 5 children or 4 children ?

there is a problem in this que
it is not mention that each one get atleast one thing , here we can take zero object for any one or not , i think we can

answer is 12-1c94-1)+11-1c(4-1)+10-1c(4-1)+9-1c(4-1)=425

15c3+14c3+13c3+12c3..this vl b the correct ans..this combination gives all possible ways

lets satisfy sana by giving her 3 chocolate first and 1 chocolate each to remaining 4 children so total choc remaining will be 8... now this 8 chocolate can be distributed among 5 children in { 8+5-1)C(5-1) = 495 ways .. now this formula (n+r-1)C(r-1) means distributing n identical things among r thing such that these r can get any number of things including 0....
which means in 495 sana may have got 0,1,2,3,4,5,6,7 or all 8 chocolates... but this is wrong as sana cannot have more than 6 chocolates so we have to eliminate the no of ways through which she got more than 6 chocolates..... that cases are when she got 8 or 7 or 6 or 5 or 4 chocolates (as she already have 3 chocolates so 3+3 is okay but 3+4i=7 is wrong)
NOW
1)she can get 8 more chocolate in only 1 case=1
2)she can get 7 more chocolate in 4C1=4 (4C1 because in this case 1 choc is remaining to distribute among 4 childs.... n=4 r=1 so 4C1)
3)she can get 6 more chocolate in 5C1=5 (because now now 2 are remaining among 4 childs.... 4+2-1C2-1)
4)she can get 5 more chocolate in 6C2=15 (because now now 3 are remaining among 4 childs.... 4+3-1C3-1)
5)she can get 4 more chocolate in 7C3=35 (because now now 4 are remaining among 4 childs.... 4+4-1C4-1)
.
.
.
total making 60 cases where she can get 6+ chocolates thus
295-(1+4+5+15+35)=295-60=235 cases
ANS:235 and it is the correct answers for sure.......

book me to 435 hai awnser
?
can someone please help me in this

arun sharma que

12-1c94-1)+11-1c(4-1)+10-1c(4-1)+9-1c(4-1)=425

12c4+11c4+10c4+9c4