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Chapter 2: Problem 5

Various six digit numbers can be formed by permuting the digits 666655 . All arrangements are equally likely. Given that a number is even, what is the probability that two fives are together?

Short Answer

Expert verified
The probability is \( \frac{4}{5} \).

Step by step solution

01

- Determine the total number of permutations

Calculate the total number of permutations of the digits 666655. Since the digits are not all unique, use the formula for permutations of multiset: \[ P = \frac{n!}{n_1!n_2!...n_k!} \] Here, we have 6 digits (n=6) where 4 are 6s and 2 are 5s: \[ P = \frac{6!}{4!2!} = \frac{720}{24 \cdot 2} = 15 \]
02

- Calculate permutations where the number is even

For a number to be even, the units digit must be 6. Fix the last digit as 6, and permute the remaining 5 digits (66655): \[ P_{even} = \frac{5!}{4!1!} = \frac{120}{24} = 5 \]
03

- Calculate the number of even permutations with two fives together

Consider the two 5s as a single unit. This gives us the digits 66(55)6. Fix the last digit as 6. We now have 4 positions to fill with the units \{66, (55)\}, giving us: \[ P_{two5s} = \frac{4!}{3!1!} = \frac{24}{6} = 4 \]
04

- Compute the probability

The probability is the ratio of favorable outcomes to total outcomes. Here, it's the number of even permutations where two 5s are together (Step 3) over the total number of even permutations (Step 2): \[ P = \frac{P_{two5s}}{P_{even}} = \frac{4}{5} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Permutations of Multiset
When dealing with permutations of a multiset, we need to consider repetitions of elements. For example, in the given exercise, the digits are 666655. When digits repeat, we use a specific formula to calculate permutations:

\[ P = \frac{n!}{n_1!n_2!...n_k!} \]

Here, we have 6 digits where 4 are 6s and 2 are 5s. Plug these into the formula:

\[ P = \frac{6!}{4!2!} \]

This equals 15. This formula ensures we don't overcount permutations with repeated elements.
Probability Calculation
To find probabilities, we calculate the ratio of favorable outcomes to the total possible outcomes. In this exercise, we first determined our total permutations for a 6-digit number from the multiset.

Then, we found how many of those permutations result in an even number by fixing the last digit as 6. Finally, we focused on the scenario where two 5s are together. The probability formula is:

\[ P = \frac{P_{two5s}}{P_{even}} \]

We computed:

\[ P = \frac{4}{5} \]

This systematic approach provides a clear path to solving such problems.
Combinatorial Analysis
Combinatorial analysis is a branch of mathematics focused on counting, arrangement, and combination of elements. In this exercise, we use factorials and arrangements to resolve counts.

When we deal with permutations and constraints, factorials (n!) are frequently used. Factorial of a number n (n!), is the product of all positive integers from 1 to n.

For example:

\[ 5! = 5 \times 4 \times 3 \times 2 \times 1 = 120 \]

Real-world applications often require breaking big problems into smaller sub-problems, like identifying favorable outcomes before total outcomes.
Even Numbers
To classify a number as even, its last digit must be 0, 2, 4, 6, or 8. In our exercise, the number must end with 6 to be even.

This constraint reduces the number of permutations. To compute permutations of an even number, we fix the last digit to 6 and then permute the rest. This means our calculations change to reflect this fixed position:

Permutations for (66655) is:

\[ \frac{5!}{4!1!} \]

Which equals 5. This step narrows the solution by only considering relevant permutations.
Permutations with Constraints
Constraints can dramatically alter permutation calculations. In our case, we have two constraints: the number is even (last digit is 6) and two 5s must be together.

To handle the second constraint, we treat two 5s as a single unit. This simplifies the problem by reducing the number of elements we need to permute.

Considering 66(55)6, where (55) is a single unit, our calculation is:

\[ \frac{4!}{3!1!} \]

Which equals 4. By understanding and applying constraints, we effectively narrow down permutations to those that meet specific conditions.

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Most popular questions from this chapter

An Einstein solid (in 3D space) has 100 lattice sites and 300 phonons, each with energy \(\hbar \omega=0.01 \mathrm{eV}\). (a) What is the entropy of the solid (give a number)? (b) What is the temperature of the, solid (give a number)?

A deck of cards contains 52 cards, divided equally among four suits, Spades (S), Clubs (C), Diamonds (D), and Hearts (H). Each suit has 13 cards which are designated: \((2,3,4,5,6,7,8,9\), \(10, \mathrm{~J}, \mathrm{Q}, \mathrm{K}, \mathrm{A}\) ). Assume that the deck is always well shuffled so it is equally likely to receive any card in the deck, when a card is dealt. (a) If a dealt hand consists of five cards, how many different hands can one be dealt (assume the cards in the hand can be received in any order)? (b) If the game is poker, what is the probability of being dealt a Royal Flush (10, J, Q, K, and A all of one suit)? (c) If one is dealt a hand with seven cards, and the first four cards are spades, what is the probability of receiving, at least one additional spade?

A system consists of \(N\) noninteracting, distinguishable two-level atoms. Each atom can exist in one of two energy states, \(E_{0}=0\) or \(E_{1}=\varepsilon\). The number of atoms in energy level, \(E_{0}\), is \(n_{0}\) and the number of atoms in energy level, \(E_{1}\), is \(n_{1}\). The internal energy of this system is \(U=n_{0} E_{0}+n_{1} E_{1}\). (a) Compute the multiplicity of microscopic states. (b) Compute the entropy of this system as a function of internal energy. (c) Compute the temperature of this system. Under what conditions can it be negative? (d) Compute the heat capacity for a fixed number of atoms, \(N\).

A fair six-sided die is thrown \(N\) times and the result of each throw is recorded. (a) If the die is thrown \(N=12\) times, what is the probability that odd numbers occur three times? If it is thrown \(N=120\) times, what is the probability that odd numbers occur 30 times? Use the binomial distribution. (b) Compute the same quantities as in part (a) but use the Gaussian distribution. (Note: For part (a) compute your answers to four places.) (c) Plot the binomial and Gaussian distributions for \(N=2\) and \(N=12\).

Consider a lattice with \(N\) spin- 1 atoms with magnetic moment \(\mu\). Each atom can be in one of three spin states, \(S_{z}=-1,0,+1\). Let \(n_{-1}, n_{0}\), and \(n_{1}\) denote the respective number of atoms in each of those spin states. Find the total entropy and the configuration which maximizes the total entropy. What is the maximum entropy? (Assume that no magnetic field is present, so all atoms have the same energy. Also assume that atoms on different lattice sites cannot be exchanged, so they are distinguishable.)
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