Green, blue, red

The number on the blue die strictly between the numbers on the red and green dice. So the $3$ numbers are distinct. Now, we can choose $3$ numbers form $1$ to $10$ in ${10 \choose 3} = 120$ ways where $3$ numbers are distinct.

Now, we can arrange any $3$ ( Distinct ) numbers in $2$ ways. They are :

$1$ : The $3$ numbers are increasing from left to right. In this arrangement we can made ${10 \choose 3} = 120$ triples.

$2$ :The $3$ numbers are decreasing from left to right. In this arrangement we can made ${10 \choose 3} = 120$ triples.

So, total we can made $2 \times 120 = \boxed { 240 }$ triples

First solution

We can count by considering each value for the blue die. Let us first assume the green die rolls higher than the red die, and we can multiply our answer by 2. For any value $b$ of the blue die, the red die can roll $1,2,\cdots,b-1$ That is $b-1$ possibilities. The green die can roll $b+1,b+2,\cdots,10$ That is $10-b$ possibilities. It is easy to find $(b-1)(10-b)$ for $b$ from 1-10, and that is $1\cdot 8+2\cdot 7+3\cdot 6+4\cdot 5+5\cdot 4+6\cdot 3+7\cdot 2+8\cdot 1=120$ Multiplying by 2, we get $\boxed{240}$ .

Alternate solution

Note that for any triple of distinct numbers from 1-10, we can arrange it in exactly 2 ways, $r<b<g$ and $g<b<r$ . There are $\dbinom{10}{3}=120$ ways to pick a triple, so the answer is $2\cdot120=\boxed{240}$ .

If we pick any $3$ distinct values that will show up on the dice, then we can order these values so that the blue die has the middle value, and then there are $2$ ways to order the red and green dice. As the order in which we pick the values does not matter, we have $2\times\binom{10}{3}=2\times120=\boxed{240}$ total ways.

Direct casework lets us check our answer. If the green die is $1$ and is smaller than the red die, then the blue die has $8$ possible outcomes; for each of these, the red die has $8,7,6,\cdots,1$ possible outcomes, respectively. This makes a total of $1+2+\cdots+7+8=36$ outcomes for this case.

Our next case is where the green die is $2$ , and it is still smaller than the red die. The same method gives $1+2+\cdots+6+7=28$ possible outcomes. As the green die gets larger, there are fewer possible values for the blue die, and thus fewer for the red. Our first case had value $1+2+\cdots+7+8=\frac{(8)(9)}{2}=\binom{9}{2}$ ; similarly, our next case has $\binom{8}{2}$ , and so on until our last case of $\binom{2}{2}$ . We multiply our sum by $2$ to account for when the green die is larger than the red die, so our total is $2\times(\binom{2}{2}+\binom{3}{2}+\cdots+\binom{9}{2})=2\times(\binom{10}{3})=2\times120=\boxed{240}$

We need to find such that in arrangements of (Green(G),Blue(B),Red(R)),no. in Green die<no. in blue dice<no.in Red die---(1) or no. in Green die>no. in blue dice>no.in Red die---(2). Clearly, for either set of arrangements, the no. of possible arrangements are equal.,i.e, no. of arrangements in (1) = no. of arrangements in (2). Hence,it is sufficient to find the no. arrangements in any one only. To do that, we select to count the no. of arrangements in (1).Then, if G=1,B=2,R=3,4,5,6,7,8,9,10 we get 8 possible arrangements.If G=1,B=3,R=4,5,...,10,we get 7 possible arrangements.Continuing this way till G=1,B=9,R=10,we get a total of (8+7+6+...+1) arrangements.Similarly, we can find the no. of arrangements for other values of G(till G=8,B=9,R=10),we notice that the total no. of such arrangements(in (1)) is:(1+2+...+8)+(1+2+...+7)+(1+2+..+6)+......(1+2)+1=120. Hence, total no. of desired arrangements=120x2=240

We focus on the blue die: (The minimum value and the maximum value of the number on the blue die is 2 and 9 respectively)

Case 1, when the number on the blue die is $2$

The number smaller than it will be $1\rightarrow1$ value

The number larger than it will be $3\cdots10\rightarrow8$ values

There are $1\times8$ triples.

Case 2, when the number on the blue die is $3$

The number smaller than it will be $1,2\rightarrow2$ values

The number larger than it will be $4\cdots10\rightarrow7$ values

There are $2\times7$ triples.

Case 3, when the number on the blue die is $4$

The number smaller than it will be $1\cdots3\rightarrow3$ values

The number larger than it will be $5\cdots10\rightarrow6$ values

There are $3\times6$ triples

Now, we seek for patterns, the number of triples are $1\times8+2\times7+\cdots+8\times1=120$

However, the combinations can occur in two cases

1)Green dice is smaller than blue dice

2)Red dice is smaller than blue dice

So, the overall number of triples are $120\times2=240$

Let g, b, r be the numbers on the green, blue and red dice respectively. Note that $g \ne r$ . WLOG, let $g > r$ . Note that if b is fixed, there are b - 1 ways to choose r, and 10 - b ways to choose g. So, the number of ways is $0 \times 9 + 1 \times 8 + 2 \times 7 + ... + 8 \times 1 + 9 \times 0 = 120$ . But, note that this is just for the case when $g > r$ . For $g < r$ , there are another $120$ more ways. This makes a total of $240$ triplets.

all 3 same = 10 any 2 same = 3(10)(9) total cases = (10)(10)(10)=1000 all different = 1000-270-10=720 blue at centre=720/3=240

We are trying to find the number of ordered triples such that the numbers in each of the triples are either in strictly ascending or strictly descending order, so each of the numbers must be distinct. There are ${10 \choose 3} = 120$ ways to choose $3$ distinct numbers from $1$ to $10$ , and there are $2$ ways to order $3$ distinct numbers in ascending or descending order, so our answer is $120*2=\fbox{240}$ .

Denote $r,b,g$ the colors, then WLOG suppose $r < g$ , we wish to find number of solutions to $r < b < g$ with $1 \leq r,b,g \leq 10$ , if $r = n$ , then there are $$1 + 2 + 3 + \dots + (9 - n) = \frac{(9 - n)(10 - n)}{2}$$ solutions. Thus the number of solutions is $$\sum_{i = 1}^{10} \frac{(9 - i)(10 - i)}{2} = 120$$ but we assumed $r < g$ , so we need to take the number twice, i.e. the number of solutions is $$2 \cdot 120 = \boxed{240}$$

Looking at the response above, you see that the problem's statement that there are 1000 different ordered triples is correct.

Now we need ordered triples in which the second number is between the first two.

Let's call the three rolls, r1, r2, and r3.

How many ordered triples are there where r1 < r2 < r3?

If the first die roll is a 1, the second is 2, the third can be 3 through 10. That is 8 ordered triples. If the first roll is a 1, the second is a 3, the third one can be 4, through 10. That is 7 more ordered triples. This goes on until you have 1, 9, 10. This gives you a total of 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1 = 36 ordered triples.

Then you go on to 2, 3, 4, followed by 2, 3, 5, etc. 7 + 6 + 5 + 4 + 3 + 2 + 1 = 28

Then starting with 3, 4, n: 6 + 5 + 4 + 3 + 2 + 1 = 21

Then 4, 5, n: 5 + 4 + 3 + 2 + 1 = 15

Then 5, 6, n: 4 + 3 + 2 + 1 = 10

Then 6, 7, n: 3 + 2 + 1 = 6

Then 7, 8, n 2 + 1 = 3

Finally 8, 9, 10: 1

When you add all these nunmbers of ordered triples up you get 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 120

Now if you add the ordered triples for which r1 > r2 > r3, you get another 120 orderred triples.

The answer is there are 240 ordered triples in which the middle number is strictly between the first and third numbers.

In other words- [1(1+1)/2 + 2(2+1)/2 + 3(3+1)/2 + 4(4+1)/2 + 5(5+1)/2 + 6(6+1)/2 + 7(7+1)/2 + 8(8+1)/2]2=240

Suppose the numbers on the green, blue, red dice are $g,b,r$ respectively.

We fix $b$ .

Suppose $g < b < r$ . The other case $g > b > r$ is exactly the same only with $g$ and $r$ swapped.

There are $b-1$ positive integers less than $b$ ; each of these can be the value of $g$ . There are $10-b$ positive integers greater than $b$ ; each of these can be the value of $r$ . So for each value of $b$ , there are $(b-1)(10-b)$ possible tuples $(g,r)$ such that $g < b < r$ .

Since we can also have $g > b > r$ , we multiply the above by $2$ . So we get $2(b-1)(10-b)$ valid tuples $(g,r)$ for each $b$ .

Finally, we just sum over all values of $b$ :

$\displaystyle\sum_{b=1}^{10} 2(b-1)(10-b)$

$= 2(0)(9) + 2(1)(8) + 2(2)(7) + \ldots + 2(9)(0)$

$= 0 + 16 + 28 + 36 + 40 + 40 + 36 + 28 + 16 + 0$

$= \boxed{240}$

Call the number showing on the green die $g$ , on the blue die $b$ and on the red die $r$ . There is an easy way to find all possible triples: just pick three distinct positive integers not larger than $10$ (there are ${10 \choose 3}$ ways to do this). Then write down these three integers in both ascending as descending order (so, $2$ ways). These triples are exactly all possible values for $(g,b,r)$ . Hence, there are ${10 \choose 3} \times 2 = \boxed{240}$ triples that satisfy the conditions.

We just have to choose 3 numbers and put them in order(increasing or decreasing). So the total is $2.10!/7!3!$ =240

We can use symmetry to solve this problem since the probability the number on the blue die is strictly between the others is the same as when the number on the red die is strictly between the others, etc

Therefore we just need to find the number of ways to get three different numbers which is $10\times9\times8=720$

We divide this by $3$ to account for the symmetry and we get $240$

Any triple of distinct numbers from 1-10 can be counted twice such that the blue die is in between the red and green dice. For instance, the triple (4, 5, 7) can satisfy the conditions for both (red, blue, green) and (green, blue, red).

Thus, the answer is 2(10C3) = 2(120) = 240.

Since the number on the blue die must be strictly between the other two, our three numbers must be distinct. There are C(10,3)=120 ways to choose 3 distinct integers among the numbers 1-10. Then we know that although the middle number must be on the blue die, the other two numbers can be ordered in two ways (1: red die shows largest numbers, 2: green die shows largest). This means that each of our triples represents a pair of possible rolls, so our final answer is 120*2=240.

If Red rolls a 1 and Green rolls a 10, the number of possibilities is 8. When Red rolls a 1 and Green rolls a 9, the number of possibilities is 7. We notice that while Red stays at 1, as Green goes down by 1, the number of possibilities goes down by 1. Therefore when Red is 1, the total possibilities is 8+7+6+5+4+3+2+1=36. Similarly, when Red is 2, the total is 7+6+5+4+3+2+1=28, so our total for Red is 36+28+21+15+10+6+3+1=120. We can also do the same for Green, so we multiply by 2, giving us 240

There are six combination of green, blue, and red die: GBR, GRB, BRG, BGR, RBG, RGB. Only 2 of these have blue die. If all the numbers of the die that are rolled are distinct, then a number will have to be between 2 others. This result is $\binom{10}{3}=120$ . 2 combinations of die have blue in between, so we multiply our result by 2 to get $\boxed{240}$

There are 10 C 3 = 120 ways to choose 3 distinct values (with red being the smallest, and then blue, and then green) from 1-10, and then you need to multiply that by 2 because the red and green can switch, so 120*2 = 240!

Clearly, none of the numbers appearing on three dices can be mutually equal as otherwise, one can not be exactly between other two. Select any three numbers out of 1-10 in 10C3 ways. Select the extremes for red and green dice in 2 ways and middle one for blue dice in 1 way. Hence, total no. of ways = 2*(10C3) = 240

Let the values of the red, blue, and green dice be $R, B, G$ respectively. We take cases based on the value of $B.$

• It is impossible for $B$ to be $1$ because then neither $R$ nor $G$ can be smaller than $B.$

• If $B = 2,$ we choose the order of the values in $2$ ways; then the smaller value has to be $1$ and the larger value has to be in the set $\{3, 4, \ldots, 10\},$ which gives $2(1)(8) = 16$ possible rolls.

• If $B = 3,$ we choose the order in $2$ ways; then there are two choices for the smaller value and $7$ choices for the larger value. This gives $2(2)(7) = 28$ possible rolls.

We work through all the other cases and add up the results to get $2(1)(8) + 2(2)(7) + 2(3)(6) + \ldots + 2(7)(2) + 2(8)(1) = \boxed{240}$ possible rolls in which the value of the blue die is strictly between the values of the other dice.

Let $a$ be the green number , $b$ the blue one and $c$ the red one. We can have either $a > b > c$ or $a < b < c$ . Let us assume WLOG $a > b > c$ .

For every value of $a$ we can have $9 - a + 1$ different values of $b$ , and hence $10 - b + 1$ values of c. Therefore , for every value of $a$ we can have ${10-a \choose 2}$ different combination, and adding them all we get ${10 \choose 3} = 120$ combinations.

Hence we have $120*2 = 240$ triples.

Any roll where a number is duplicated clearly doesn't lead do a solution.

However, for every possible unordered combination of three different numbers, two solutions are apparent (sorting the numbers ascending, and sorting descending).

So the answer is $2 * {}^{10}C_{3}$ , which is $240$

Set "b" for blue , , "r" for red and "g" for green . case I : r<b<g fix b= 2 , r has only one choice and g has eight choices to choose from 3, 4 ,..... 10 number of triplets = 8 fix b = 3 , r has two choices 1 and 2 . g has seven choices 4, 5,... 10 . numbers of triplets = 14 fix b = 4 we will get number of triplets = 18, fix b= 5 , we will get number of triplets = 20 fix b= 6 , we will get number of triplets = 20 fix b= 7 , we will get number of triplets = 18 fix b= 8 , we will get number of triplets = 14 fix b= 9 , we will get number of triplets = 8

total number of triplets = 120. similarly when g<b<r , we shal get total number of triplets = 120. In all there will be 240 triplets where blue lies between red and green

number on the blue between the numbers on the red and green if number on the blue 1 and 10, there is not anything way if number on the blue 2 and 9 there are 1 8 2=16 2=32 ways if number on the blue 3 and 8 there are 2 7 2=28 2=56 ways if number on the blue 4 and 7 there are 3 6 2=36 2=72 ways if number on the blue 5 and 6 there are 4 5 2=40 2=80 ways So the total is 240

Let x, y, z = the value shown in the blue, green, and red dice respectively if x = 1, there is no values for y and z such that the condition will be satisfied. if x = 2, we can have 1 value for y and 8 values for z, a total of 8 combinations. if x = 3, we have 2 for y and 7 for z for a total of 14 combinations. if x = 4, we have 3 for y and 6 for z, a total of 18 combinations. if x = 5, we have 4 for y and 5 for z for a total of 20 combinations. so far, we have 60 combinations. it can be easily shown that these combinations is the same for 5<x<=10. thus we now have 60 2 = 120 combinations. also, it does not matter if y>z or z>y, so the values of the two can be interchanged for any combination. thus, we have 120 2 = 240 combinations all in all.

The numbers on the three dice must be distinct, so there are ${10 \choose 3} = 120$ ways to choose the three numbers. Once the numbers are selected, there are 2 ways for the dice to properly show these numbers. Either the red must show the largest number and the green the smallest, or vice versa (blue showing the middle number). This yields $120\cdot 2 = 240$ total triples.

let x = required solution no. of turns when red and blue dice show same number and green shows a different number=10x9=90 no. of turns when all dice show same number =10 => 3x+3C2*90+10=1000 => x=240!

Let the value rolled on the red, blue, and green dice be $r, b, g$ , respectively. Then for each admissible triple, either $r < b < g$ or $g < b < r$ . Suppose the former; then for each $2 \le b \le 9$ , there are $b - 1$ choices for $r$ and $10 - b$ choices for $g$ , and therefore $\sum_{b = 2}^9 (b-1)(10-b) = 120$ admissible triples with $r < b < g$ . But by symmetry, there are also $120$ triples with $g < b < r$ , for a total of $\boxed{240}$ such triples.