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14 May 2026, 19:00
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
56% (01:51) correct
44%
(02:04)
wrong
based on 103
sessions
History
Date
Time
Result
Not Attempted Yet
In how many ways can 10 identical chocolates be distributed among three children such that each child gets at least one chocolate?
A. 36
B. 66
C. 72
D. 78
E. 84
A. 36
B. 66
C. 72
D. 78
E. 84
ankushsambare
Joined: 13 Jun 2022
Last visit: 04 Sep 2026
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Schools: ISB '27 Wharton '26 INSEAD Aug '26
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14 May 2026, 22:40
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We need the number of positive integer solutions to:
x + y + z = 10
where each child gets at least one chocolate, so:
x ≥ 1, y ≥ 1, z ≥ 1
Using the stars and bars formula for positive integer solutions:
Number of ways = C(n − 1, r − 1)
Here:
n = 10 chocolates
r = 3 children
So:
C(9, 2) = 9 × 8 / (2 × 1) = 36
Answer: A. 36
x + y + z = 10
where each child gets at least one chocolate, so:
x ≥ 1, y ≥ 1, z ≥ 1
Using the stars and bars formula for positive integer solutions:
Number of ways = C(n − 1, r − 1)
Here:
n = 10 chocolates
r = 3 children
So:
C(9, 2) = 9 × 8 / (2 × 1) = 36
Answer: A. 36
BongBideshini
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15 May 2026, 10:02
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First, give 1 chocolate to each child.
So:
10 − 3 = 7 chocolates remain.
Now distribute these 7 identical chocolates freely among 3 children.
Visualize 7 chocolates as zeros:
0 0 0 0 0 0 0
We need 2 bars to divide them among 3 children.
0 0 | 0 0 0 | 0 0
Now total objects:
The bars can be placed anywhere, including together or at the ends.
Total combination
9C2
= 9! / (2! * 7!)
= 36
Answer: A
So:
10 − 3 = 7 chocolates remain.
Now distribute these 7 identical chocolates freely among 3 children.
Visualize 7 chocolates as zeros:
0 0 0 0 0 0 0
We need 2 bars to divide them among 3 children.
0 0 | 0 0 0 | 0 0
Now total objects:
- 7 zeros
- 2 bars
The bars can be placed anywhere, including together or at the ends.
Total combination
9C2
= 9! / (2! * 7!)
= 36
Answer: A
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