Problem of the Week

Problem 15 Solution

Congratulations to Harry Auld from Parkside Community College, Keo Osman Simon Balle School, Hertford, Felix Reiter from Bottisham Village College, for successfully solving this problem!

i) When selecting the first card to place down in front of you, the magician has 13 options. After selecting one card, 12 remain. This means that for the selection of the second card, there are 12 options, and so on. The total number of options is the product:

$13 \times 12 \times \cdots \times 1$

This is known as $13!$ (the $!$ denoting factorial).

ii) In the second case, order does \textbf{not} matter. Initially, we want to select 7 cards from the 13, which can be done in:

$13 \times 12 \times \cdots \times 7$

ways. However, this counts some selections multiple times. For example, selecting the 1st card then the 2nd card is counted separately from selecting the 2nd card and then the 1st, even though they result in the same set of cards. To avoid this double-counting, we divide by the number of ways to arrange the 7 selected cards, which is $7!$ . This removes the duplicates, resulting in the number of combinations:

$\frac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7}{7!}$

This simplifies to:

$\frac{13 \times 12 \times 11 \times 10 \times 9 \times 8 \times 7}{7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 1716$

iii) The first part is $n!$ , since you have $n$ options for the first card, $n-1$ for the second, and so on until you have only 1 option left.

The second part is a bit trickier. Similar to the previous case, we are selecting items, so factorials are helpful. However, we aren’t multiplying all the whole numbers from $n$ down to 1; instead, we are multiplying all the whole numbers from $n$ down to $n-k+1$ (the $k$ numbers we are choosing). This gives the expression:

$\frac{n!}{(n-k)!}$

We now need to divide by the number of ways to arrange the $k$ selected items to avoid counting the same set of items multiple times. Since there are $k!$ ways to arrange the $k$ items, we divide by $k!$ . This simplifies to:

$\frac{n!}{k!(n-k)!}$

Problem of the Week

Problem 15 Solution

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