Probability-related questions

1. Find the number of different sums that can be obtained by using one, some or all of the numbers in the set {2^0,2^1,2^2...,2^n}.

2. Find the number six-digit numbers that can be formed using the digits from the number 112233. If these numbers are arranged in ascending order, find the 30th number and the median.

3. How many ways can six people be divided into three groups of two people each?

Please show me the working for these kind of questions, thank you very much.

PS: I always wonder, for question 2, did it really related to probability?

Re: Probability-related questions

Quote:

Originally Posted by

**alexander9408** 1. Find the number of different sums that can be obtained by using one, some or all of the numbers in the set {2^0,2^1,2^2...,2^n}.

Exactly what do you mean by "sums". In "United States English" we mean specifically "adding" while "British English" encompasses all kinds of calculations. What operations are you allowed to use?

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2. Find the number six-digit numbers that can be formed using the digits from the number 112233. If these numbers are arranged in ascending order, find the 30th number and the median.

. There are 6 choices for the first digit, 5 choices for the second digit, 4 for the third, etc.

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3. How many ways can six people be divided into three groups of two people each?

There are 6 choices for the first person to put into the first group, 5 choices for the second person to put into the first group, 4 choices for the first person to put into the second group etc. The difference between this and problem two is that this problem say "three groups"- it does NOT specify a 'first group', 'second group', 'third group'. So you have to **divide** by the number of ways of "ordering" the three groups- there are three ways to choose a "first group", two ways to choose a "second group"- and then of course, there the last group is fixed.

Quote:

Please show me the working for these kind of questions, thank you very much.

PS: I always wonder, for question 2, did it really related to probability?

Re: Probability-related questions

Hello, alexander9408!

Quote:

1. Find the number of different sums that can be obtained by using one, some or all

of the numbers in the set $\displaystyle \{2^0,\,2^1,\,2^2\,\hdots\.2^n\}.$

This is the number of possible $\displaystyle (n\!+\!1)$-digit binary numbers.

The number is: .$\displaystyle 2^{n+1}-1$

Re: Probability-related questions

Quote:

Originally Posted by

**HallsofIvy** Exactly what do you mean by "sums". In "United States English" we mean specifically "adding" while "British English" encompasses all kinds of calculations. What operations are you allowed to use?

. There are 6 choices for the first digit, 5 choices for the second digit, 4 for the third, etc.

There are 6 choices for the first person to put into the first group, 5 choices for the second person to put into the first group, 4 choices for the first person to put into the second group etc. The difference between this and problem two is that this problem say "three groups"- it does NOT specify a 'first group', 'second group', 'third group'. So you have to **divide** by the number of ways of "ordering" the three groups- there are three ways to choose a "first group", two ways to choose a "second group"- and then of course, there the last group is fixed.

I don't get the second one, mind to explain more, please?

As for the third one, is that (6x5x4x3x2x1)/3!2!2!2! ?

Quote:

Originally Posted by

**Soroban** Hello, alexander9408!

This is the number of possible $\displaystyle (n\!+\!1)$-digit binary numbers.

The number is: .$\displaystyle 2^{n+1}-1$

Sir, what is binary number?

Re: Probability-related questions

Quote:

Originally Posted by

**alexander9408** Sir, what is binary number?

You never did explain what is meant by *a sum* in this question.

In this case, a binary number is same an base 2 representation of a positive integer (that explains the -1)

Quote:

Originally Posted by

**alexander9408** I don't get the second one, mind to explain more, please?

As for the third one, is that (6x5x4x3x2x1)/3!2!2!2! ?

You have the third one correct.

There $\displaystyle \frac{6!}{2^3}$ ways to rearrange the string $\displaystyle 112233$.