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05 Oct 2021, 00:15
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
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There are 5 adults and 3 children in a meeting. If 2 adults and 2 children are randomly chosen to give a speech one by one, how many different arrangements of speeches are possible?
A. 30
B. 120
C. 360
D. 480
E. 720
A. 30
B. 120
C. 360
D. 480
E. 720
05 Oct 2021, 01:56
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There are 5 adults and 3 children in a meeting. If 2 adults and 2 children are randomly chosen to give a speech one by one, how many different arrangements of speeches are possible?
No of ways you can select 2 adults out of 5 = 5C2 = \(\frac{5*4}{2!}\)= 10
No of ways you can select 2 children out of 3 = 3C2 = \(\frac{3*2}{2!} \)= 3
So the total no of ways you can select 2 adults and 2 children in the meeting = 5C2 * 3C2
For every 4 selected people, no of different orders they can give the speech = 4!
For Example:
if A1,A2,C1,C2 are selected out of 4 adults A1,A2,A3,A4 and 3 children C1,C2,C3.
No of different orders where A1,A2,C1,C2 can give the speech is 4! = 24 as the speech is given one by one.
Writing down the possible speech orders below.
1. A1,A2,C1,C2
2. A1,A2,C2,C1
3. A1, C1,A2,C2
4. A1, C1,C2,A2
5. A1,C2,A2,C1
6. A1,C2,C1,A2
..
..
..
..
..
..
..
24. C2,C1,A2,A1
So the total no of different arrangements of speeches possible for all combinations = 5C2 * 3C2 * 4! = 10* 3 * 24 = 720
Option E is the correct answer.
Thanks,
Clifin J Francis,
GMAT SME
No of ways you can select 2 adults out of 5 = 5C2 = \(\frac{5*4}{2!}\)= 10
No of ways you can select 2 children out of 3 = 3C2 = \(\frac{3*2}{2!} \)= 3
So the total no of ways you can select 2 adults and 2 children in the meeting = 5C2 * 3C2
For every 4 selected people, no of different orders they can give the speech = 4!
For Example:
if A1,A2,C1,C2 are selected out of 4 adults A1,A2,A3,A4 and 3 children C1,C2,C3.
No of different orders where A1,A2,C1,C2 can give the speech is 4! = 24 as the speech is given one by one.
Writing down the possible speech orders below.
1. A1,A2,C1,C2
2. A1,A2,C2,C1
3. A1, C1,A2,C2
4. A1, C1,C2,A2
5. A1,C2,A2,C1
6. A1,C2,C1,A2
..
..
..
..
..
..
..
24. C2,C1,A2,A1
So the total no of different arrangements of speeches possible for all combinations = 5C2 * 3C2 * 4! = 10* 3 * 24 = 720
Option E is the correct answer.
Thanks,
Clifin J Francis,
GMAT SME
21 Feb 2025, 03:49
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The Trap here is that each adult and child is a unique element.
We have 5 As and 3 Cs and we need 2 As and 2 Cs for the Speech. (Example combination: AACC)
This can be done in 5c2 * 3c2 ways = 10 * 3 ways = 30
Now, each adult and child is a unique element! Therefore order does matter.
Therefore the number of ways to pick an order for 2 As and 2 Cs = 30 * 4! = 720
Ans E
If the order didn't matter, we would have 30 * 4!/2!.2! = 180 ways. It is not in the answer choices but expect that in the real exam
We have 5 As and 3 Cs and we need 2 As and 2 Cs for the Speech. (Example combination: AACC)
This can be done in 5c2 * 3c2 ways = 10 * 3 ways = 30
Now, each adult and child is a unique element! Therefore order does matter.
Therefore the number of ways to pick an order for 2 As and 2 Cs = 30 * 4! = 720
Ans E
If the order didn't matter, we would have 30 * 4!/2!.2! = 180 ways. It is not in the answer choices but expect that in the real exam
