Question 1021867: I have total 3 things and 5 slots . How can i fill this 5 empty slots using only 3 things ? please answer me . thanks in advance
Answer by mathmate(429)   (Show Source):
You can put this solution on YOUR website!
Question:
I have total 3 things and 5 slots . How can i fill this 5 empty slots using only 3 things ? please answer me . thanks in advance
Solution:
It is obvious that there will have to be at least two empty slots, by the pigeon hole principle.
The question is probably how many ways I can fill three different objects into 5 empty slots, where order of the slots is important (permutation), but the order of putting objects in the slots is not important (combinations).
Note that there is no limitation on how many objects can be put in one slot.
We further assume that all three objects must end up in one slot or another (i.e. all put in the slots).
So putting three objects in the same slot, there are 5 ways.
|ABC| | | | |
| |ABC| | | |
... etc
| | | | |ABC|
5 ways..................(1)
Putting two objects in one slot and one in one of the remaining slots:
|AB|C| | | |
|AB| |C| | |
...
|AB| | | |C|
that makes 4 ways.
However, there are 4 ways for each position of AB, from 1 to 5. That makes 4*5=20 ways.
In addition, instead of AB, it could have been AC or BC, so that makes 4*5*3=60 ways..........(2)
Now if we put the objects into individual slots, the first object has 5 choices, second 4 and third 3 for a total of 60 ways.............(3)
Adding up (1), (2) and (3) we have 5+60+60=125 ways.
Additional Question:
here is the question : (Q) How many ways can we arrange the letters in the word PROVIDE so that no two vowels are adjacent? please give me the solution with explanation . thanks
Solution:
To solve this problem, we understand that
1. When there are no restrictions, the number of arrangements is 7!=5040.
This number includes cases where two of the vowels are together, or three vowels are together.
2. When there are at least two vowels together, we can treat this as a word of 6 letters, giving 6!=720 arrangements. However, since there are 6 ways of choosing the two letters (order counts), there are 6*6!=4320 cases where at least two vowels are together. NOTE that this case (2) includes cases where all three vowels are together!
3. When all three vowels are together, we can treat the case as a word of 5 letters, with 5!=120 arrangements. However, there are 3!=6 ways (order counts) of choosing the 3 letters, so there are 3!*5!=720 arrangements.
To sum up, if we don't want any of the vowels together, we have
# of arrangements
= case (1)-( (2)-(3) )
= case (1) -(2) + (3)
= 5040 - 4320 + 720
= 1440
Looking it another way:
The four consonants can be arranged in 4!=24 ways.
The vowels can be "inserted" around the consonants in 5C3=10 ways, since the order of the vowels counts, we multiply by 3!=6. The total number of ways is therefore 24*10*6=1440 ways as before.
|C|C|C|C|
vowels can be inserted in any of the five "|" positions, but not two in the same slot, so 5C3=5!/(3!2!)=10.
Since ordering of vowels is important, we have {EIO,EOI,IEO,IOE,OEI,OIE}
for a total of 6 ways = 3!.
In counting problems, it is always advisable to attempt counting in different ways to verify the answers against each other.
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