Shrutiejha wrote:
Hey,
Can anyone go through the questions below and clear my doubt as I have been trying to understand it but can't find a logical answer anywhere.
In how many ways can a person send invitation cards to 6 of his friends if he has four servants to distribute the cards?
(a) 6^4
(b) 4^6
(c) 24
(d) 120
Ans - Each invitation card can be sent in 4 ways. Thus,4×4×4×4×4×4=4^6.
My Doubt - Why can't it be 6^4 , as in the other way around?
In how many ways can 5 prizes be distributed to 8 students if each student can get any number of prizes?
(a) 40
(c) 8^5
(b) 5^8
(d) 120
Official Answer - 8^5
Doubt - Again, why can't it be 5^8?
Thanks a lot for your help!
First question:
The first helper can send an invitation to any 6 friends. However, the second helper can't send to all 6 since one friend already has received the invitation.
Also, the first helper can send invites to mote than one friend. Hence, we cannot assume that the 1st helper has 6 options and the 2nd has 5 options etc.
When you put a ball in a box, the options are for the balls, not the boxes. Similarly, here the options are for the friends - each can receive invites from any of the 4 helpers
Question 2:
Each prize can be assigned to any of the 8 students. Thus, first prize has 8 options, 2nd prize has 8 options and so on: 8*8*8*8*8 = 8^5 (the same student can get all 5 prizes)
If we try the reverse logic:
First student can get any 5 prizes
But what about the second student? It depends on how many the first got!
Hope this helps
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