Permutations: 3-digit numbers formed using 1-7 if....

[fred2028]

So here's my math question.

1. How many 3-digit numbers can be formed using only the numbers 1-7 if the number 2 must be included?

So I did 7P2. Because 2 must be in the 3-digit number set, we really only have the option of choosing 2 digits.

7P2 = 42
42 * 3 = 126

I then multiplied it by 3 since the number 2 can be in the first, second, or third digit of the number.

However, the answer in the back of the sheet says it's 210. The only way to get 210 is to do 7P3, but that doesn't make sense, since you are not choosing all 3 digits. Anyone can explain this?

 

[pka]

There is one important question: "Can a digit be repeated?"

 

[fred2028]

There is one important question: "Can a digit be repeated?"

That is a good question. I have a screenshot of my worksheet.

http://i9.tinypic.com/2nbhqop.jpg

 

[pka]

The answer I get if you allow repartition: 3(7)<SUP>2</SUP>.
Three place to place the 2 and 7 choices for the other two.

I think that the [127] is a typo it should be [147].

 

[fred2028]

The answer I get if you allow repartition: 3(7)<SUP>2</SUP>.
Three place to place the 2 and 7 choices for the other two.

[127] would be the answer.
And 3(7)^2 is 147, isn't it?

 

[pka]

I think that it is a typo!

 

[fred2028]

I think that it is a typo!

I guess so ...

This unit is so confusing. I mean the textbook questions are OK, but this worksheet is so hard.

 

[pka]

I have checked #2 & #3 and the answers given are correct.

 

[fred2028]

I have checked #2 & #3 and the answers given are correct.

Number 2 I got correct, but number 3:
In my understanding, the way to get to the answer is:

11! - 2 = 39,916,798

11! because there are 11 factorial ways to place all players on a bench. However, Joey and Jill don't wanna be together, so Joey:Jill and Jill:Joey are gone, so I subtracted 2.

But my answer does not agree with the answer on the PDF ...

 

[pka]

\(\displaystyle \L\left( {\begin{array}{c}
{10} \\
2 \\
\end{array}} \right)\left( {9!} \right)\left( {2!} \right).\)

 

[fred2028]

\(\displaystyle \L\left( {\begin{array}{c}
{10} \\
2 \\
\end{array}} \right)\left( {9!} \right)\left( {2!} \right).\)

I don't really understand where you got the numbers from...?

 

[pka]

_^_^_^_^_^_^_^_^_^_
Ten places to put the two boys; 9! ways to arrange the others; 2! to arrange the boys.

 

[Tuugii]

I guess:

there are three main types of this 3-digit number cintaining the digit-2 at least one time: 2xx , x2x ,and xx2 .

Each of them has ways: 1*7*7 , 7*1*7 ,and 7*7*1 }=>thus, (7^2)*3

Tuugii

 

[fred2028]

Thank you everyone! The teacher walked us through this question and the last question, and told us that they weren't what he expected, so he started us off and I got the right answer.

Tuugii's thought process is what he showed us.

 

[pka]

Tuugii's thought process is what he showed us.

If that is true, I would like to know how he explained #3.

 

[fred2028]

Tuugii's thought process is what he showed us.

If that is true, I would like to know how he explained #3.

He didn't explain every single one ...
He said this is an assignment we are handing in, so he's not going to "take it up".

 

[Tuugii]

My apologies to all, I guess my previous solution( 3*(7^2) ) is completely wrong....

I have over counted by 20 numbers, the real answer is 127.

solution:
(1) this is clear that there are 3 types of 3-digit numbers containing the digit-2 at least one time: 2xx, x2x, xx2
(2) and lets divide all 3-digit numbers into two sections: {200-299} and {100-199 and 300-999}
(3) all numbers of the type 2xx are in the interval {200-299}, thus it is clear that these are: 1*7*7=1*7^2=49;
(4) the numbers of the type x2x are: 6*1*7=42 ; this is because there are 6 hundreds without 200 containing the numbers {1-7}: {100, 300, 400, 500, 600, 700} and 7digits {1-7};
(5) the numbers of the type xx2 are: 6*6*1=6^2*1=36 ; this is because there are 6 hundreds and 7 tens{10, 20, 30, 40, 50, 60, 70}, but we have counted all numbers with {20} in the type x2x, thus there are 6 tens to count.
(6) and the result is: 49+42+36=127

Tuugii