Arranging distinct color balls inside an equilateral triangle box.

combinationscombinatoricscontest-mathpermutationssolution-verification

Ten balls of the same shape and size but having different colors are
to be arranged into an equilateral triangular box as shown. Find the
number of ways in which this arrangement can be done

enter image description here

My solution approach :-
Number of ways in which we can select the center ball = $10$

Number of ways in which we can select the balls around the center ball = $^9C_6=84$
Number of ways in which we can arrange the $6$ balls around the center ball = $(6-1)!=5!=120$ as they will be in circular arrangement and we know that the circular arrangement of $n$ distinct items = $(n-1)!$
Total number of ways of arranging $6$ balls around the center ball = $84 \times 120=10080$

Number of ways of arranging the remaining $3$ balls around the corners = $(3-1)!=2!=2$

Hence the number of ways should be = $10 \times 10080 \times 2= 201600$

But this is the wrong answer. What am I doing wrong? Please help me on this !!!

Thanks in advance !!!

Best Answer

Method 1

Choose 4 for the bottom row, and arrange them, in $10C4\times4!$ ways.

Choose 3 for the next row, and arrange them, in $6C3\times3!$ ways.

Choose 2 for the next row, and arrange them, in $3C2\times2!$ ways.

Place the final ball at the top in $1$ way.

Multiply these together to get $3628800$. But now divide by $3$ because we can rotate the whole image thus duplicating arrangements already found by a factor of $3$.

Final answer: $1209600$

Method 2

Choose the top ball in $10$ ways.

Choose and arrange the next row down in $9C2\times2!$ ways.

Choose and arrange the third row in $7C3\times3!$ ways.

The last four balls can be arranged at the bottom in $4!$ ways.

Multiply all these together and divide by $3$ as before and get the same answer: $1209600$.

Method 3

Choose the central ball in $10$ ways.

Choose the surrounding $6$ balls and arrange them in $9C6\times6!$ ways. This is not the same as a circular table arrangement: the rotations are dealt with at the end.

The last $3$ balls can be arranged in $3!$ ways.

Multiply all these together and divide by 3 for the same reason as before, and get the same answer as before: $1209600$.

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