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2 questions:

  1. How many odd $14$ digit numbers can you compose of ten $1$'s and four $2$'s such that in between every pair of $2$'s there are at least two $1$'s?

Here I could have $4$ or $5$ spaces for the $1$'s to be in between, is the maximum $5$ enough? So $C(5+10-3, 5-1)$?

  1. In how many ways can you arrange $5$ red balls, $5$ yellow balls and $2$ blue balls (where balls of the same color are considered identical) such that the ball on the right edge will be red and there are not two red balls next to each other?

Total ways to arrange the balls: $C(12,3)$.

2 options that the ball on the right is not red. so far $C(12,3)-2$. I'm not sure whats the right to account for red balls not being next to each other.

Thanks,

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  • $\begingroup$ After clicking on "ASK QUESTION" you are supposed to ask one question, not two questions. If you want to ask another question, you should click "ASK QUESTION" again. $\endgroup$ – bof Dec 1 '14 at 11:54
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For the first question: The number of ways a $14$ digit number consisting of ten $1$'s and four $2$'s can be composed is the number of ways ten $1$'s can be placed in a list of four $2$'s. Since there are five spaces in which the $1$'s can be placed, this is equivalent to the number of solutions of the equation

$$x_1 + x_2 + x_3 + x_4 + x_5 = 10$$

in which $x_k$ represents the number of $1$'s in the $k$th space.

Since the $14$ digit number is odd, $x_5 \geq 1$. Since there must be at least two $1$'s between successive $2$'s, $x_2 \geq 2$, $x_3 \geq 2$, and $x_4 \geq 2$. Hence, the number of $14$ digit odd numbers consisting of ten $1$'s and four $2$'s in which at least two $1$'s appear between successive $2$'s is the number of solutions in the integers of the equation

$$x_1 + x_2 + x_3 + x_4 + x_5 = 10$$

subject to the restrictions \begin{align*} x_1 & \geq 0\\ x_2 & \geq 2\\ x_3 & \geq 2\\ x_4 & \geq 2\\ x_5 & \geq 1\\ \end{align*}

Let \begin{align*} y_1 & = x_1\\ y_2 & = x_2 - 2\\ y_3 & = x_3 - 2\\ y_4 & = x_4 - 2\\ y_5 & = x_5 - 1 \end{align*}

Then the number of $14$ digit odd numbers consisting of ten $1$'s and four $2$'s in which at least two $1$'s appear between successive $2$'s is the number of solutions in the non-negative integers of the equation

\begin{align*} y_1 + y_2 + 2 + y_3 + 2 + y_4 + 2 + y_5 + 1 & = 10\\ y_1 + y_2 + y_3 + y_4 + y_5 & = 3 \end{align*}

which is the number of ways in which four addition signs can be placed in a list of three $1$'s, which is

$$\binom{3 + 4}{4} = \binom{7}{4} = 35$$

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Ok I don't think this is the conventional way to do it but here goes anyway feel free to critique my answer...

  1. You know that the last digit is 1, thus the first 13 digits have 4 2's and 9 1's.

There has to be at least 2 1's between each pair of 2's, so the configuration requires the following groups of 1's as a minimum - 0,2,2,2,0. (5 gaps because there are 4 twos - 3 gaps between the twos, and 2 gaps between the outermost twos and the outside.

Take away the 6 1's that are required to be in between the 2's, and you are left with 3 1's that can be placed in each of the five gaps.

How many ways can you place the three ones into five gaps?

The answer is 7C4 = 35

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