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08 Feb 2022, 10:32
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
56% (01:56) correct
44%
(02:09)
wrong
based on 264
sessions
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In how many different ways could a schoolteacher distribute 5 identical apples to three children, if each child could receive any number of apples from 0 to 5?
A) 9
B) 15
C) 21
D) 25
E) 30
A) 9
B) 15
C) 21
D) 25
E) 30
09 Feb 2022, 01:29
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Number of ways of dividing 'n' identical objects into 'r' groups such that each group can contain any number of objects is given by (n+r-1)C(r-1)
We have n=5 and r=3. Using the formula, we get 7!/(2!*5!)=21 C)
Similar question: https://gmatclub.com/forum/in-how-many- ... 06279.html
We have n=5 and r=3. Using the formula, we get 7!/(2!*5!)=21 C)
Similar question: https://gmatclub.com/forum/in-how-many- ... 06279.html
General Discussion
06 Dec 2025, 05:34
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I think with this question you just have to look at different situations.
first situation: 5-0-0 which can be arranged in 3!/2! ways = 3
second situation: 4-1-0 which can be arranged in 3! ways = 6
third situation: 3-2-0 which can be arranged in 3! ways = 6
fourth situation 2-2-1 which can be arranged in 3!/2! ways = 3
fifth situation 3-1-1 which can be arranged in 3!/2! ways = 3
final answer: 6+6+3+3+3=21
first situation: 5-0-0 which can be arranged in 3!/2! ways = 3
second situation: 4-1-0 which can be arranged in 3! ways = 6
third situation: 3-2-0 which can be arranged in 3! ways = 6
fourth situation 2-2-1 which can be arranged in 3!/2! ways = 3
fifth situation 3-1-1 which can be arranged in 3!/2! ways = 3
final answer: 6+6+3+3+3=21
nislam
