Combinatorics : AMC 8 2008 Problem 14

Combinatorics is a field of Mathematics where we study in how many ways we can arrange some number of given objects following some certain rules of arrangements.

Three $\text{A's}$, three $\text{B's}$, and three $\text{C's}$ are placed in the nine spaces so that each row and column contain one of each letter. If $\text{A}$ is placed in the upper left corner, how many arrangements are possible?

$\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 6$

AMC 8 , 2008 Problem 14

Combinatoric

6 out of 10

Mathematical Circles

Knowledge Graph

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Use some hints

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Lets solve the problem together !!!

lets name all the small boxes : We will call the box on the intersection of $i^{th}$ row and $j^{th}$ column $\text{Box}(i,j)$

$1$ of the three A's is fixed in the $\text{Box}(1,1)$

Then A can not occur in the $1^{st}$ row and $1^{st}$ column.

So to place a B in the $1^{st}$ row we have 2 choices[ $\text{Box}(1,2)$ and $\text{Box}(1,3)$ ] and then only $1$ choice left to place C in the $1^{st}$ row.

Now can you think about the placements of A, B and C is the second row ??

A can not occur in the $1^{st}$ column so there are $2$ choices to place A in $2^{nd}$ row and then only one choice left to place C in $2^{nd}$ row.

Now think about the $3^{rd}$ row.

After placing all the letters in the described manner we are automatically left with only one way to place A, B and C in the third row.

So the total number of choice in $1^{st}$ row is $2$ for each of these ways there are $2$ choices for the $2^{nd}$ row and for each of these $2\times 2$ choices there is only one choice for the $3^{rd}$ row.

Hence the total number of choices : $2 \times 2 \times 1=4$.

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