Find the total number of arrangements of 4 cards from 7 cards

Question

The 7 cards are arranged in a random order in a straight line.

A A A B B R R

4 of the 7 cards are arranged in a random order in a straight line Find the total number of arrangements of these 4 cards

Answer 1

You have the multiset: $\{A\cdot 3, B\cdot 2, R\cdot 2\}$ and are looking for permutations of 4 elements. So, you can have the following:

$$\{A\cdot 3, B\cdot 1\},\{A\cdot 3, R\cdot 1\},\{A\cdot 2, B\cdot 2\},\{A\cdot 2, B\cdot 1,R\cdot 1\},\{A\cdot 2, R\cdot 2\},\{A\cdot 1, B\cdot 2,R\cdot 1\},\{A\cdot 1, B\cdot 1,R\cdot 2\},\{B\cdot 2, R\cdot 2\}$$

This gives the total number of 4-permutations: $$\dfrac{4!}{3!1!}+\dfrac{4!}{3!1!}+\dfrac{4!}{2!2!}+\dfrac{4!}{2!1!1!}+\dfrac{4!}{2!2!}+\dfrac{4!}{1!2!1!}+\dfrac{4!}{1!1!2!}+\dfrac{4!}{2!2!} = 4+4+6+12+6+12+12+6=62$$

This is the same as Thomas Andrews found.

Answer 2

You can compute it by brute force. Let $p_m(a_1,a_2,a_3)$ be the number of ways of arranging $m$ cards of three types, with $a_i$ cards of card type $i.$

You are trying to compute: $p_4(3,2,2).$ Note, the order of the $a_i$ doesn't change the value.

Then you get:

$$\begin{align}p_4(3,2,2)&=p_3(2,2,2)+p_3(3,1,2)+p_3(3,2,1)\\&=p_3(2,2,2)+2p_3(3,2,1) \end{align}$$

Then you have $$\begin{align}p_3(2,2,2)&=3p_2(2,2,1)\\p_3(3,2,1)&=p_2(2,2,1)+p_2(3,1,1)+p_2(3,2,0)\end{align}$$ So we get:

$$p_4(3,2,2)=5p_2(2,2,1)+2p_2(3,1,1)+2p_2(3,2,0).$$

Now, $$\begin{align}p_2(2,2,1)&=2p_1(2,1,1)+p_1(2,2,0)\\&=2\cdot 3 + 2\\&=8\\ p_2(3,1,1)&=p_1(2,1,1)+2p_1(3,1,0)\\&=3+2\cdot 2\\&=7\\ p_2(3,2,0)&=p_1(2,2,0)+p_1(3,1,0)\\&=2+2\\&=4 \end{align}$$

So $$p_4(3,2,2)=5\cdot 8 + 2\cdot 7 + 2\cdot 4= 62$$