A trailer can be loaded with 1 car, or with 4 cars.
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The problem said between 1 and 4, so it could only have 2 or 3, so shouldn't the answer also be 280?
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Sorry but what do you mean by 2 or 3 and the answer also 2 8 0 ?
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Since the number of cars per trailer has to be between 1 and 4, only 2 cars or 3 cars per trailer are allowed. There are 280 ways to do this.
It is not "strictly between 1 and 4", so you are allowed to have 1 car or 4 cars.
I'd add in an explanation for clarity.
yeah the answer is 280
banglaralobd.com
this question of which standerd
same as my approach except for the division by 2 to take care of the symmetry (which was my mistake). In the first case, i suppose we could have calculated the same as 8C3 * 6C2 / 2.. Also, aren't there 3 ways in which the 3 , 3 and 2 can be arranged (3, 3 and 2 / 2, 3 and 3 and 3, 2 and 3) thus requiring division by 3
what's that operator (8 2) that u have used in first step to get 280??
This is equivalent to the number of 8 letter words consisting of A , B , C where each letter appears between 1 and 4 times - by writing down to which trailer each car goes, and then dividing by 3 ! , since the trailers are identical.
We will do it using exponential generating functions: for each letter, f X ( x ) = x + 2 ! x 2 + 3 ! x 3 + 4 ! x 4 - for X = A , B , C . So the generating function is f ( x ) = f A ( x ) ⋅ f B ( x ) ⋅ f C ( x ) = ( x + 2 ! x 2 + 3 ! x 3 + 4 ! x 4 ) 3 And we are looking for the coefficient of 8 ! x 8 , which is 4 6 2 0 - by direct computation while remembering that we only need x 8 .
So, the number of 8 letter words as described above is 4 6 2 0 , and so the answer will be 4 6 2 0 / 3 ! = 7 7 0
Can you please explain the concept of exponential g . f. ?
1)
(
4
8
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∗
(
3
4
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∗
(
1
1
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=
2
8
0
(4 3 1)
2)
(
4
8
)
∗
(
2
4
)
∗
(
2
2
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=
4
2
0
(4 2 2)
3)
(
3
8
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∗
(
3
5
)
∗
(
2
2
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=
5
6
0
(3 3 2)
Total number of ways = ( 1 ) + 2 ( 2 ) + ( 3 ) = 7 7 0 .
Can you explain why "Total = ( 1 ) + 2 ( 2 ) + ( 3 ) ? Many people feel that it should be ( 1 ) + ( 2 ) + ( 3 ) for a total of 1260.
Note: To type latex, you just need to add the brackets \ ( and \ ) with spaced removed. You can see how I edited your solution, as an example.
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In case (2) you are counting the train with 2 cars twice and in case (3) you are counting twice the train with 3 cars. Because of that, you have to divide by 2.
i require some more clear explanation can some help in getting out of this jugulary????
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Motivation When we see such combinatorics question with a maximum number of items being placed into a box, we usually think of using the maximum, or m a x idea as shown below:
Note The 3 trailers are identical.
If the m a x of the number of cars in the trailers is 3 , we will have only 1 case to consider:
The number of ways to arrange the 8 cars in this order is simply: ( 2 8 ) ⋅ 2 ( 3 6 ) = 2 8 0 . We choose 2 cars out of 8 to put into 1 trailer, then we simply choose 3 out of the remaining 6 cars, then divide by 2 because choosing 3 cars c 1 , c 2 , c 3 to put into one trailer is the same as choosing c 1 , c 2 , c 3 to put into the other trailer.
If the m a x is 4 , there are 2 cases to consider:
( 4 , 3 , 1 ) : Number of ways to arrange = ( 4 8 ) ⋅ ( 3 4 ) = 2 8 0
( 4 , 2 , 2 ) : Number of ways to arrange ( 4 8 ) ⋅ 2 ( 2 4 ) = 2 1 0 .
Therefore, the total number of ways is 2 8 0 + 2 8 0 + 2 1 0 = 7 7 0 . We are done. :)