I have this question where I get
I want the length of $48$ inches. How would the $2$ lego be combined to get $48$ inches? The equation is $$ 4a + 6b = 48 $$ which has the solutions
So it was $5$ difference way you can achieve $48$ inches Now the question is, in how many ways can you line the lego ?
HINT:
In the case $a=0, b=8$ you have only one way (if the lego pieces are identical), think about it as putting eight letter bs in a row $$ bbbbbbbb $$ can be done only one way as long as the lego are identical. If not then you can say that you can choose from $8$ different one for the first place, $7$ for the second and so on giving $1\cdot 2\cdot\ldots\cdot 8=8!$
In the case $a=3, b=6$ you have to arrange three as and six bs, the question is in how many different words you can create using these letters $$ aaabbbbbb $$ is one of them. Once again we have to ask whether these are as and bs are indentical in between themselves. If they are all different you get $9!$ otherwise you can swap the first and the second a to get $$ aaabbbbbb. $$ In fact all the as and bs are swappable and you have to sort of remove these, by dividing by all the different ways of ordering them giving you $$ \frac{9!}{3!\cdot 6!} $$ where $3!$ is all the ways to order the as and $6!$ is all the ways to order the bs.
Hope this helped
You can further read up on permutation of a multiset
This is very awkwardly stated! In the first place, we don't have "two legos", we have two kinds of legos! The problem is to find how many of each kind we can use.
Then you are seeking non-negative integers, a and b, such that 4a+ 6b= 48. The first thing I would do is divide through by 2: 2a+ 3b= 24. Now "2" and "3" are prime so, by the Euclidean algortihm, we know there must exist integers (not necessarily positive), a and b, such that 2a+ 3b= 1. In fact, that's obvious- 3- 2= 1. Now multiply by 24: 3(24)- 2(24)= 2(-24)+ 3(24)= 24. That is, a= -24, b= 24 is a solution. But then, a= -24+ 3k, b= 24- 2k is also a solution for every integer, k: 2(-24+ 3k)+ 3(24- 2k)= -48+ 6k+ 72- 6k= 24.
We want non-negative integers so we need to choose k so that both a= -24+ 3k and b= 24- 2k are positive. 3 divides into 24 8 times so [tex]k\ge 8[/tex] will make a non-negative. 2 divides into 24 12 times so [tex]k\le 12[/tex] will make b non-negatifve.
The solutions are
s, not legos). Now, for each of your cases you found, count the number of orderings of the corresponding number of 4 inch lego and 6 inch lego using binomial coefficients. For instance, with three $4$ inch lego and six $6$ inch lego, you have nine spaces total and so $\binom{9}{3}$ ways to choose which of the spaces are used by the $4$ inch lego. – JMoravitz Mar 09 '20 at 12:21