# No Of ways of arranging plates belonging to two different sets?

There are $$n$$ triangular plates on which numbers from $$1$$ to $$n$$ are written and there are $$m$$ circular plates on each of them number from $$1$$ to $$m$$ are written. We have to find the total no ways of arranging these taken all at time such that triangular and circular plates are in order respectively.

E.g., there are $$3$$ circular plates named as $$c_1, c_2, c_3$$ also there are $$4$$ triangular plates named as $$t_1, t_2, t_3, t_4$$ One of the possible arrangements of plates is

$$t_1~c_1~c_2~t_2~c_3~t_3~t_4$$

$$t_1, t_2, t_3, t_4$$ are in order; $$c_1, c_2, c_3$$ are in order respectively.

• Welcome to MathSE. When you pose a question here, it is expected that you include your own thoughts on the problem. Explain what you know, show what you have attempted, and explain where you are stuck. This tutorial explains how to typeset mathematics on this site. – N. F. Taussig Dec 15 '18 at 12:20

Without loss of generality let $$n\ge m$$. Then we can order triangular plates first and then place circular plates in-between: $$* \ t_1 \ * \ t_2 \ * t_3 \ * * \ * t_{m-1} \ * \ t_m \ * t_{m+1} \ * \ * \ * t_n$$ $$c_1$$ can be immediately before or after $$t_1$$, $$c_2$$ can be immediately before or after $$t_2$$ and so on $$c_m$$ can be immediately before or after $$t_m$$. So, there are $$2^m$$ ways to order them.

For example, for your case $$n=3$$, $$m=4$$, there are $$2^3=8$$ ways: $$c_1t_1 \ c_2t_2 \ \color{red}{c_3t_3} \ t_4;\\ c_1t_1 \ c_2t_2 \ \color{red}{t_3c_3} \ t_4;\\ c_1t_1 \ \color{red}{t_2c_2} \ c_3t_3 \ t_4;\\ \color{red}{t_1c_1} \ c_2t_2 \ c_3t_3 \ t_4;\\ c_1t_1 \ \color{red}{t_2c_2 \ t_3c_3} \ t_4;\\ \color{red}{t_1c_1} \ c_2t_2 \ \color{red}{t_3c_3} \ t_4;\\ \color{red}{c_1t_1 \ c_2t_2} \ c_3t_3 \ t_4;\\ \color{red}{c_1t_1 \ c_2t_2 \ t_3c_3} \ t_4;\\$$

• Your answer is incorrect. There are $10$ admissible arrangements for two triangular and three circular plates. – N. F. Taussig Dec 15 '18 at 12:01
• @N.S.Taussig, how come? For two triangular and three circular, there are only $4$ arrangements: $t_1c_1t_2c_2c_3 \\ t_1c_1c_2t_2c_3 \\ c_1t_1t_2c_2c_3 \\ c_1t_1c_2t_2c_3$. – farruhota Dec 15 '18 at 12:06
• Read my answer. It seems you are making an additional assumption about the relative order of the plates. – N. F. Taussig Dec 15 '18 at 12:07
• I see now. Indeed, the ordering of the plates is not required. I guess the example given by OP distracted. Oh well, it looks I answered a different question then. Cheers! – farruhota Dec 15 '18 at 12:12

Hint: Since the numbers on the plates of each type must appear in increasing numerical order, an arrangement is completely determined by choosing which $$n$$ of the $$n + m$$ positions required for $$n$$ triangular plates and $$m$$ circular plates are occupied by the triangular plates.

For example, consider the case of three circular plates and two triangular plates. The ten admissible arrangements are:

$$c_1c_2c_3t_1t_2$$

$$c_1c_2t_1c_3t_2$$

$$c_1c_2t_1t_2c_3$$

$$c_1t_1c_2c_3t_2$$

$$c_1t_1c_2t_2c_3$$

$$c_1t_1t_2c_2c_3$$

$$t_1c_1c_2c_3t_2$$

$$t_1c_1c_2t_2c_3$$

$$t_1c_1t_2c_2c_3$$

$$t_1t_2c_1c_2c_3$$