Possible Lego People

Put all the Lego pieces into a pile, and create Lego people one by one as follows.

Choose one head, one body and one set of legs to make the first person. There are four heads, four bodies, and four sets of legs in our pile, so there are $4 \times 4 \times 4 = 4^3$ potential people.

Repeat the process. There are now three heads, three bodies and three sets of legs in the pile so there are $3 \times 3 \times 3 = 3^3$ options for the second Lego person.

Similarly, there are $2 \times 2 \times 2 = 2^3$ options for the third Lego person. Finally there is just one head, one body and one set of legs left in our pile, so there is just one choice for the final Lego person.

So there are $4^3 \times 3^3 \times 2^3 \times 1$ sequences of Lego people. But we are asked for the number of different sets of four Lego people, i.e the order doesn't matter.

There are $4! = 4 \times 3 \times 2 \times 1$ ways of ordering four Lego people, so the number of different sets of four Lego people is $\frac{4^3 \times 3^3 \times 2^3 \times 1}{4 \times 3 \times 2 \times 1} = 4^2 \times 3^2 \times 2^2 = 576$ .

The first Lego man have 4 \times 4 \times 4 = 4^3 combinations, the second 3^3 , the third 2^3 and the last 1^3. To simplify, the number of ordered sets of 4 Lego people = 4! \cdot 4! \cdot 4!. But since the order of the Lego people doesn't matter, the number of sets = \frac {4! \cdot 4! \cdot 4!}{4!} = 576

the first person can choose legs in 4 ways, torso in 4 ways and head also in 4 ways = 64 ways (4x4x4)

the second person can choose legs in 3 ways, torso in 3 ways and head also in 3 ways = 27 ways (3x3x3)

the third person can choose legs in 2 ways, torso in 2 ways and head also in 2 ways = 8 ways (2x2x2)

the fourth person can choose legs in 1 ways, torso in 1 ways and head also in 1 ways = 1 ways (1x1x1)

therefore total ways = 64 x 27 x 8 x 1

However, we are not interested in arrangement, i.e. anyone can be 1st, 2nd, 3rd or 4th. Therefore divide the entire expression by 4! (read 4 factorial)

Hence the answer is 64 x 27 x 8 x 1 / 4! = 576

To simplify calculations, we first assume that order matters.

There are $4\times4\times4$ possibilities for the first Lego person, and similarly, there are $3\times3\times3$ possibilities for the second, $2\times2\times2$ for the third, and $1\times1\times1$ for the fourth.

Thus there are $4\times4\times4\times3\times3\times3\times2\times2\times2\times1\times1\times1 = (4\times3\times2\times1)^3$ different permutations, where order matters.

However, in this problem, order does not matter, thus we have to divide $(4\times3\times2\times1)^3$ by $4!$ . Thus there are $576$ different sets of Lego people.

Como existem 4 pessoas e cada pessoa é composta de três partes, então existem 4 cabeças, 4 troncos e 4 pernas distintas.

Se fixarmos as 4 cabeças, podemos permutar os 4 troncos com as cabeças, isto é, 4! = 24 possibilidades.

Agora, basta colocar os pés. Temos 4 pés que podem ser novamente permutados, isto é, 4! = 24 possibilidades.

Como os eventos são independentes, usamos o Princípio Fundamental Multiplicativo. Assim, o número de pessoas é 24.24 = 576 .

Miles' four Lego people implies four different legs, torsoes, and heads. Thus, there exists $4 \times 4 \times 4$ possibilities for the first lego person,, leaving $3 \times 3 \times 3$ for the second, $2 \times 2 \times 2$ for the third, and finally 1 possibility for the final lego person, giving $(4!)^3$ total possibilities of sets of 4 lego people. For some distinct set of lego people combinations, a combination of lego people can exist at the first, second, third, or fourth person, leaving another combination for the next three, another for the next two, and finally one for the last. This gives $4*3*2*1$ similar sets for every distinct case or $4!$ . Dividing the total number of sets by the number of similar sets, $(4!)^3/4!$ , yields $(4!)^2$ absolutely distinct sets, or 576 different sets.

we have 4 heads, 4 set of legs and 4 torsos. we set the 4 heads. For these heads, we can rearrange the torsos as following: For the first head we may use one of the 4 torsos, for the second one, we may use one of three, for the third one, we may use 2 and at last, for the fourth one, we can use only one. That is 4!. For each rearrangement of head and torso as a only one piece, we proceed as before, i.e., as we did using the heads and the torsos. In this way, we have 4! for each previous rearrangement. Therefore, we have at the end (4!)(4!) different sets of 4 Lego people.

(4!)(4!) = 576.

Solution 1

There are $4! = 24$ ways to assign the torsos to the heads. There are $4! = 24$ ways to assign the legs to the combined head and torso. So in total there are $24 \times 24 = 576$ ways to assign put the Lego people back together.

Solution 2

To create the first Lego person, there are 4 possibilities for each of the head, torso, and legs, for a total of $4^3 = 64$ possibilities. For the second lego preson there will be $3^3 = 27$ possibilities. For the third person, there will be $2^3 = 8$ possibilities. For the last person there is only a single possibility. To calculate the total number of possibilities, we multiply these values together to get $64 \times 27 \times 8 = 13824$ . However, the order does not matter, so we have to divide through by the number of different ways we could construct the same set of Lego people ( $4! = 24$ ). So the answer is $\frac{13824}{24} = 576$ .

There are 4 lego people with a head, a torso, and a set of legs each. Hence, we can write it in the form

4 4! 3! = 576

** 3! here because there are 3 parts consisting head, torso, and sets of legs.

The first person can choose legs in 4 ways, torso in 4 ways and head also in ways --> 64 ways the second person 3x3x3--- > 27 ways the third person 2x2x2 ---> 8 ways the fourth person 1x1x1 ---> 1 ways

therefore total ways = 64 x 27 x 8 x 1 However, we are not interested in arrangement, i.e. anyone can be 1st, 2nd, 3rd or 4th. Therefore divide the entire expression by 4!

Hence the answer is 64 x 27 x 8 x 1 / 4 = 576

There are 4! ways to connect legs with torso. Now there are another 4! ways to connect with these legs connected to torso. Thus answer is $(4!)^2 = 576$

the first person can choose legs in 4 ways, torso in 4 ways and head also in ways --> I.e 4x4x4=64 ways the second person 3x3x3=27 ways the third person 2x2x2 = 8 ways the fourth person 1x1x1 = 1 ways

So total ways are = 64 x 27 x 8 x 1 But we are not interested in arrangement, i.e. anyone can be 1st, 2nd, 3rd or 4th. Therefore divide the entire expression by 4!

Hence the answer is (64 x 27 x 8 x 1) / 4! = 576