Please help with this counting problem: 16 pebbles are arranged in a 4 by 4 array. Find the number of way of choosing four pebbles, so that no two of the chosen pebbles lie in the same row or column.
+26367 Please help with this counting problem: 16 pebbles are arranged in a 4 by 4 array.
Find the number of way of choosing four pebbles,
so that no two of the chosen pebbles lie in the same row or column.
My attempt:
16 choices for a first pebble,
9 choices for a second pebble, and
4 choices for a third pebble, and
1 choice for a fourth pebble gives a total of \(16\times9\times4\times1 = 576\), if the pebbles are numbered,
when not, gives a total of \(\dfrac{16\times9\times4\times1}{4!} = 24\)
By rotation there are 2 different possibilities:
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By rotation there are 4 different possibilities:
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