Latest Elitmus Aptitude Question SOLUTION: Zada has to distribute 15 choclates among 5 of her children Sana,ada,jaya,amir and farhan. She has to make sure that sana gets at least 3 and ada at most 6 choclates. In how many w

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(#M40152550) ELITMUS question Quants Keep an EYE

Zada has to distribute 15 choclates among 5 of her children Sana,ada,jaya,amir and farhan. She has to make sure that sana gets at least 3 and ada at most 6 choclates. In how many ways can this be done?
A.495 B.77 C. 417 D. 435

Asked In Elitmus ayanava (8 years ago)

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Tamal Chakraborty  (8 years ago)

Actual Ans is : 425

1st slot : 4 choco are distributed. remains 15-4=11
case 1 when san get 3 chocolate: ( 8+4-1)C(4-1) = 11C3 = 165

case 2 when sen get 4 chlt 10C3 = 120

case3 sen get 5 chlt 9C3 = 84

case4 sen get 6 chklt 8C3 = 56

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

Is this solution Helpfull? Yes (2) | No (1)  

avijit sharma  (8 years ago)

This qn is present in arun sharma --> Permutation & combination --> qn. 64 lod 1

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santosh  (8 years ago)

D)435
check

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ayanava  (8 years ago)

Solve it Santosh

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avijit sharma  (8 years ago)

Does this qn came in ur elitmus exam ???

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jitendra  (8 years ago)

ans 435
check

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jitendra  (8 years ago)

495
(3,1,11)=165
(4,1,10)=120
(5,1,9)=84
(6,1,8)=56

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Vishaka Garg  (8 years ago)

@avijitsharma... derz no sch ques in arun sharma quant apti.......can u plz explain it??

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avijit sharma  (8 years ago)

vishaka garg -- yes..it is there .... on page 583...anyways...here is the solution:

We will first give 3 chocolates to Sana and 1 each
to the other 4. So we have already distributed -
chocolates. Now we are left with 8 chocolates that
have to be distributed among 5 people. So number
of ways possible is 12!/8!4! -- 495. Out of these
we have to eliminate cases in which Sana gets
more than 6 chocolates. As we have already given
her 3 chocolates that means we have to eliminate
cases in which she gets 7 or 8 or 9 or l0 or
11 chocolates out of 495 cases. She would get 11
chocolates in one case, l0 chocolates in 4C1 cases,
9 chocolates in 5C1, cases, 8 chocolates in 6C2 cases
and 7 chocolates in 7C3, cases. So, total number of
cases that need to be eliminated
is 60. So the required
number of ways is 495 - 60 =
435

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