Question

A standard deck of cards has 52 cards divided into 4 suits, each of which has 13 cards. Two of the suits ($\heartsuit$ and $\diamondsuit$, called 'hearts' and 'diamonds') are red, the other two ($\spadesuit$ and $\clubsuit$, called 'spades' and 'clubs') are black. The cards in the deck are placed in random order (usually by a process called 'shuffling'). In how many ways can we pick two different cards

Answer 1

Answer:

The number of ways to select 2 cards from 52 cards without replacement is 1326.

The number of ways to select 2 cards from 52 cards in case the order is important is 2652.

Step-by-step explanation:

Combinations is a mathematical procedure to compute the number of ways in which k items can be selected from n different items without replacement and  irrespective of the order.

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

Permutation is a mathematical procedure to determine the number of arrangements of k items from n different items respective of the order of arrangement.

$^{n}P_{k}=\frac{n!}{(n-k)!}$

In this case we need to select two different cards from a pack of 52 cards.

• Two cards are selected without replacement:

Compute the number of ways to select 2 cards from 52 cards without replacement as follows:

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

${52\choose 2}=\frac{52!}{2!(52-2)!}$

      $=\frac{52\times 51\times 50!}{2!\times50!}\\=1326$

Thus, the number of ways to select 2 cards from 52 cards without replacement is 1326.

• Two cards are selected and the order matters.

Compute the number of ways to select 2 cards from 52 cards in case the order is important as follows:

$^{n}P_{k}=\frac{n!}{(n-k)!}$

$^{52}P_{2}=\frac{52!}{(52-2)!}$

       $=\frac{52\times 51\times 52!}{50!}$

       $=52\times 51\\=2652$

Thus, the number of ways to select 2 cards from 52 cards in case the order is important is 2652.