A trailer can be loaded with 1 car, or with 4 cars.

780
770
750
760

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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?

Albert Xu
- 7 years, 3 months ago

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Sorry but what do you mean by $2$ or $3$ and the answer also $280$ ?

Happy Melodies
- 7 years, 3 months ago

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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.

Albert Xu
- 7 years, 3 months ago

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

Darren Huang
- 7 years, 3 months ago

banglaralobd.com

Masud Milon
- 7 years, 3 months ago

this question of which standerd

Haji Khalil
- 7 years, 3 months ago

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

Sundar R
- 7 years, 3 months ago

what's that operator (8 2) that u have used in first step to get 280??

Ak Sharma
- 7 years, 2 months ago

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, $f_X(x)=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}$ - for $X=A,B,C$ . So the generating function is $f(x)=f_A(x)\cdot f_B(x)\cdot f_C(x)=\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\right)^3$ And we are looking for the coefficient of $\frac{x^8}{8!}$ , which is $4620$ - by direct computation while remembering that we only need $x^8$ .

So, the number of $8$ letter words as described above is $4620$ , and so the answer will be $4620/3!=\boxed{770}$

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Can you please explain the concept of
**
exponential g . f.
**
?

A Brilliant Member
- 7 years, 3 months ago

1)
${8 \choose 4}*{4\choose 3}*{1 \choose 1} = 280$
(4 3 1)

2)
${8 \choose 4}*{4\choose 2}*{2 \choose 2} = 420$
(4 2 2)

3)
${8 \choose 3}*{5 \choose 3}*{2\choose 2} = 560$
(3 3 2)

Total number of ways $= (1) + \frac{ (2) +(3)} {2} = \boxed{770}$ .

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Can you explain why "Total $= (1) + \frac{ (2) + (3) } {2}$ ? 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.

Igor dos Santos Silva
- 7 years, 3 months ago

i require some more clear explanation can some help in getting out of this jugulary????

Rahul Sharma
- 7 years, 3 months ago

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MotivationWhen we see such combinatorics question with a maximum number of items being placed into a box, we usually think of using the maximum, or $max$ idea as shown below:NoteThe $3$ trailers are identical.If the $max$ 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: $\dbinom{8}{2} \cdot \dfrac{\binom{6}{3}}{2}= 280$ . 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 $max$ is $4$ , there are $2$ cases to consider:

$(4,3,1)$ : Number of ways to arrange $= \dbinom{8}{4} \cdot \dbinom{4}{3} = 280$

$(4,2,2)$ : Number of ways to arrange $\dbinom{8}{4} \cdot \dfrac{\binom{4}{2}}{2}= 210$ .

Therefore, the total number of ways is $280 +280+210 = \boxed{770}$ . We are done. :)