Question

What is the probability that the 2 and 3 of clubs are next to each other in a shuffled deck? Hint: Imagine the two cards accidentally stuck together and shuffled as one card.

Short answer

The probability is \(\frac{1}{26}\).

Step-by-step solution

- Total Number of Arrangements

Calculate the total number of ways to arrange the deck of 52 cards. This is given by the factorial of 52, \(52!\).

- Treat 2 and 3 of Clubs as a Single Card

Imagine the 2 and 3 of clubs are stuck together, forming one 'super card'. This reduces the number of cards from 52 to 51.

- Number of Arrangements with the Super Card

Calculate the number of ways to arrange this reduced deck of 51 cards, which is \(51!\).

- Arrangements of the Two Cards Within the Pair

Since the 'super card' can be either '2 of clubs, 3 of clubs' or '3 of clubs, 2 of clubs', there are 2 ways to arrange these two cards.

- Total Number of Favorable Arrangements

Multiply the number of arrangements of the 51 cards by the 2 arrangements of the paired cards: \(51! \times 2\).

- Calculate the Probability

Divide the number of favorable arrangements by the total number of arrangements: \[ \text{Probability} = \frac{51! \times 2}{52!} = \frac{2}{52} = \frac{1}{26} \].

Key concepts

Factorial

Understanding factorials is crucial for solving many combinatorial problems. The factorial of a number, denoted as \(n!\), is the product of all positive integers up to that number. For example, \(5!\) (read as 'five factorial') is calculated as follows: \(5! = 5 \times 4 \times 3 \times 2 \times 1 = 120\).
Factorials grow very quickly as the number increases. For instance, \(10!\) is 3,628,800. In combinatorial arrangements, we often calculate factorials for large numbers. In the given exercise, we deal with \(52!\) and \(51!\). These are extremely large numbers and usually handled using mathematical software or a calculator.
If you're solving a problem involving permutations or combinations, knowing how to compute a factorial is essential. Remember that \(0! = 1\) by definition. This special case often comes up in combinatorial formulas.

Combinatorial Arrangements

Combinatorial arrangements refer to the different ways to order or arrange a set of objects. If we're arranging all 52 cards in a deck, we calculate the total number of possible arrangements using \(52!\).
Sometimes, we need to arrange a subset or treat certain objects as a single entity, as seen in the exercise where the 2 and 3 of clubs are treated as one 'super card.' This changes our total count from 52 to 51, reducing the arrangement calculation to \(51!\).
Each object in our arrangement can impact the total number of combinations. If you have a constraint, like a set of objects needing to stay together, you can adjust your calculations accordingly. For instance, in our exercise, once the 2 and 3 of clubs are together as one super card, the total ways to arrange these 51 entities must be considered, along with the 2 ways to arrange the pair within the super card.

Card Probability

Probability in card games often involves figuring out favorable outcomes over possible outcomes. In our exercise, the problem is finding the likelihood that 2 and 3 of clubs stay next to each other after shuffling.
Using combinatorial arrangements, we first determine the total number of all possible arrangements for 52 cards, which is \(52!\). Next, by treating the 2 and 3 of clubs as a single element, we reduce the complexity to arranging 51 cards with one 'super card,' leading to \(51!\) arrangements.
Since the 'super card' can be 2 different internal arrangements ('2 of clubs, 3 of clubs' or '3 of clubs, 2 of clubs'), we multiply \(51!\) by 2. Thus, our total favorable outcomes are \(51! \times 2\).
Finally, we calculate the probability by dividing the favorable outcomes by the total outcomes: \(\frac{51! \times 2}{52!} = \frac{1}{26}\). Hence, the chance of the 2 and 3 of clubs being next to each other in a shuffled deck is \(\frac{1}{26}\).