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20 Feb 2020, 00:14
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A king has 50 identical diamond rings, which he stores in five different coloured chests. In how many ways can the king store the rings among the chests such that every chest contains at least one ring?
a) \(49C4\)
b) \(50C4\)
c) \(49C5\)
d) \(50C5\)
e) \(50C3\)
a) \(49C4\)
b) \(50C4\)
c) \(49C5\)
d) \(50C5\)
e) \(50C3\)
20 Feb 2020, 00:15
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CaptainLevi
If no restriction of atleast 1 ring..
This question is best done by taking all 50 items in a line and then taking four partitions so as to have 5 groups.
***...*l**ll*****..*l*(here the first one has >4, second one 2, third one 0, fourth again>6 and fifth just 1.) or
llll*****.....(here the first one, the second one, the third one and the fourth have 0 each and fifth all 50.)
Now we have total of 50 rings and 4 partitions so 50+4=54 out of which 4 partitions are to be chosen....\(54C4\)
If restriction of atleast 1 ring as in this question..
First distribute 1 each to 5 chests, so remaining =50-5 or 45 rings.
*l***l**...**l*l*****, so AT LEAST one between each partition.
Now take remaining 45 items in a line and four partitions so as to have 5 groups. Now we have total of 45 rings and 4 partitions so 45+4=49 out of which 4 partitions are to be chosen....\(49C4\)
A
arpitkansal, I have posted the question again as the original posted has been deleted by someone while I was responding to your query.
General Discussion
20 Feb 2020, 00:25
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A king has 50 identical diamond rings, which he stores in five different coloured chests. In how many ways can the king store the rings among the chests such that every chest contains at least one ring?
Think of the diamond rings as D
We have 50 IDENTICAL D's
We need to divide in 5 DIFFERENT boxes (chests)
Since each box must have at least 1 diamond ring, we first assign onenringnto each box (since the rings are identical, it doesn't matter which rings are chosen or in which order they are chosen)
This, there are now 50 - 5 = 45 rings left
To do the distribution, imagine that we have written down the 45 D's in a line and have put 4 partitions (denoted by P) somewhere in between those. For example:
DDD P DDDDD P DD P D P DDDD...D
=> This means one box has 3 rings, one has 5 rings, one has 2 rings, one has 1 ring and the rest go to the last box
Again: P P P P DDDDD...DDD
=> This means 4 boxes have 0 rings, and all 46 rings go to the last box
[Note: we have already assigned one ring to each box]
Thus, different positions of the Ps result in different distributions. The total number of such positions is simply the arrangement of 45 Ds and 4 Ps in a line (total 49 things)
Number of ways
= \(49!/(45! * 4!) =\) \(49C4\)
Answer A
Posted from my mobile device
Think of the diamond rings as D
We have 50 IDENTICAL D's
We need to divide in 5 DIFFERENT boxes (chests)
Since each box must have at least 1 diamond ring, we first assign onenringnto each box (since the rings are identical, it doesn't matter which rings are chosen or in which order they are chosen)
This, there are now 50 - 5 = 45 rings left
To do the distribution, imagine that we have written down the 45 D's in a line and have put 4 partitions (denoted by P) somewhere in between those. For example:
DDD P DDDDD P DD P D P DDDD...D
=> This means one box has 3 rings, one has 5 rings, one has 2 rings, one has 1 ring and the rest go to the last box
Again: P P P P DDDDD...DDD
=> This means 4 boxes have 0 rings, and all 46 rings go to the last box
[Note: we have already assigned one ring to each box]
Thus, different positions of the Ps result in different distributions. The total number of such positions is simply the arrangement of 45 Ds and 4 Ps in a line (total 49 things)
Number of ways
= \(49!/(45! * 4!) =\) \(49C4\)
Answer A
Posted from my mobile device
