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04 Mar 2026, 21:56
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B
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:
45% (02:08) correct
55%
(01:43)
wrong
based on 44
sessions
History
Date
Time
Result
Not Attempted Yet
Jimmy has 5 different flavored candies and 4 different jars. In how many ways can he put the candies in the jars such that no jar is left empty and all the candies are put in the jars?
A. 720
B. 240
C. 120
D. 9
E. 5
A. 720
B. 240
C. 120
D. 9
E. 5
05 Mar 2026, 03:15
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This is a Constrained Distribution problem — distributing distinct objects into distinct containers with no container left empty. It's trickier than a standard Combinations question because the constraint forces you to think about partitioning first.
The common trap: jumping straight to 4^5 = 1,024 (assigning each candy to any of 4 jars freely) without applying the "no jar empty" constraint. That gives 1,024 — none of the answer choices — and sends students into a spiral.
Step 1: Figure out the only valid partition.
We have 5 candies going into 4 jars with no jar empty. The only way to distribute 5 distinct items into 4 non-empty groups is 2 + 1 + 1 + 1 (one jar gets 2 candies, the other three each get 1). There is no other valid partition — 3+1+1+0 leaves a jar empty, and 2+2+1+0 does too.
Step 2: Choose which 2 candies share a jar.
C(5, 2) = 10 ways
Step 3: Now we have 4 groups — one pair and three singles. Assign these 4 groups to the 4 distinct jars.
4! = 24 ways
Step 4: Multiply.
Total = 10 × 24 = 240
Answer: B (240)
The key insight is that with 5 items and 4 non-empty jars, the partition 2+1+1+1 is forced — there's only one structure to consider. Recognizing this immediately cuts the problem from "complex counting" to a clean two-step calculation.
Takeaway: In constrained distribution problems, always identify the valid partition(s) first — it tells you exactly how many cases to calculate and prevents over-counting.
The common trap: jumping straight to 4^5 = 1,024 (assigning each candy to any of 4 jars freely) without applying the "no jar empty" constraint. That gives 1,024 — none of the answer choices — and sends students into a spiral.
Step 1: Figure out the only valid partition.
We have 5 candies going into 4 jars with no jar empty. The only way to distribute 5 distinct items into 4 non-empty groups is 2 + 1 + 1 + 1 (one jar gets 2 candies, the other three each get 1). There is no other valid partition — 3+1+1+0 leaves a jar empty, and 2+2+1+0 does too.
Step 2: Choose which 2 candies share a jar.
C(5, 2) = 10 ways
Step 3: Now we have 4 groups — one pair and three singles. Assign these 4 groups to the 4 distinct jars.
4! = 24 ways
Step 4: Multiply.
Total = 10 × 24 = 240
Answer: B (240)
The key insight is that with 5 items and 4 non-empty jars, the partition 2+1+1+1 is forced — there's only one structure to consider. Recognizing this immediately cuts the problem from "complex counting" to a clean two-step calculation.
Takeaway: In constrained distribution problems, always identify the valid partition(s) first — it tells you exactly how many cases to calculate and prevents over-counting.
