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05 Sep 2019, 04:25
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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:
43% (01:05) correct
57%
(01:13)
wrong
based on 95
sessions
History
Date
Time
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In how many of ways can 5 balls be placed in 4 tins if any number of balls can be placed in any tin?
(A) 5\(C\)4
(B) 5\(P\)4
(C) \(5^4\)
(D) \(4^5\)
(E) \(5^5\)
(A) 5\(C\)4
(B) 5\(P\)4
(C) \(5^4\)
(D) \(4^5\)
(E) \(5^5\)
Source: Nova GMAT
08 Sep 2019, 02:42
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If we have 4 destinations and 5 balls to arrange, shouldn't it be 5*5*5*5?
saratsuman
Joined: 19 Jun 2019
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09 Sep 2019, 09:29
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nehasomani33
If we have 4 destinations and 5 balls to arrange, shouldn't it be 5*5*5*5?
Here's what I can understand from the given stem.
"We have 5 balls that are to be placed into 4 tins and any number of balls can be placed in a tin"
Therefore, we have 5 balls [ let's name them B1 B2 B3 B4 B5] and 4 tins [T1 T2 T3 T4]
No of ways B1 can be placed into a tin = 4 [T1 T2 T3 T4];
No of ways B2 can be placed into a tin = 4 [T1 T2 T3 T4];
No of ways B3 can be placed into a tin = 4 [T1 T2 T3 T4];
No of ways B4 can be placed into a tin = 4 [T1 T2 T3 T4];
No of ways B5 can be placed into a tin = 4 [T1 T2 T3 T4];
Therefore, we can place 5 balls into 4 tins in 4*4*4*4*4 ways => 4^5 ways.
Hope it's clear.
