Six billiard balls, which are numbered as 2,3,5,7,8,9 inclusively, will be put into the cylindrical transparent boxes whose capacities are 3 and must be fulfilled.
Furthermore, the balls must be in descending order in the boxes.
In how many different 3-digit numbers can we generate such as 289 or 357? Could you explain the answer as explaining to a stupid?
On
Once the balls are put into the box, there is only ONE way to satisfy the order rule of decending.
For example, if ball 2,8,9 are in the box, there is only one decending order 9,8,2.
Therefore, we just need to take care how to select 3 out of these 6 to be assigned to the box : $C^6_3$ .
On
Once you choose which balls will go in a box, there is only one way to arrange them, so the answer is just the number of ways to put $3$ balls in box. Furthermore, once we choose which balls go in box A, the other $3$ must go into box B, so there is only one choice to make -- the $3$ balls that go into box $A$.
The answer is $$\binom63=\frac{6\cdot5\cdot4}{3\cdot2\cdot1}=20$$
To see this, note that we have $6$ choices for the first ball, then $5$ for the second, then $4$ for the third. This counts each combination $3!=6$ times, because choosing $abc$ is the same as choosing $acb$ or $bca$ etc. There are $3!$ ways to arrange $3$ objects, so we have overcounted by a factor of $6$, and must divide by $6$ to get the answer.
On
Let us first take a simpler case to help understand better , let's say we have 4 billiards balls numbered 1,2,3,4 and a single cylinder of capacity of 2 balls .
Now we want to put the balls in descending order , and find the possible number of numbers that can be formed .
=>1.)Let is first , calculate the total number of selections of balls that can be made , without any restrictions of putting them in descending order or anything like that .
The possible cases are : $(1,2),(1,3),(1,4),(2,3),(2,4),(3,4)$
Which are , 6 cases .
=>2.) Now let's count the number of cases , with the restriction i.e, the numbers must be in descending order .
The cases are:$(4,3),(4,2),(4,1),(3,2),(3,1),(2,1)$
that is ,also 6 cases .
As you might have noticed , the number of cases in (1) and (2) are equal , that is ,6 .
Hence we can also say that , the number of cases of putting the balls in descending order , is actually also equal to the number of cases of different selections of balls without any restriction at all!
Therefore , the number of different numbers is just , $C^6_3 * C^3_3$
This type of question is like what is the number of arrangements of the following equality in the set $1,2,3,4,5,6,7$ such that $x_1<x_2<x_3$. The answer of the question is $C(7,3)$. If it were $x_1<x_2<x_3<x_4<x_5$ ,then the answer would be $C(7,5)$.The key point is the number of subsets.
Then ,we can apply it this question such that $C(6,3)$ for the first tube and $C(3,3)$ for the second. Hence , $C(6,3) \times C(3,3)=20$
NOTE:When we find the number of subsets , then you dont need to worry about the ordering in the tubes.They will come in there in the wanted order