1) consider the number of grid paths from the orgin in a coordinate graph to each of the following points:
a) (4,7) b) (3,7) c) (4,6)
2) simplify: (n-r)! / (n-r+1)!
3) solve (n+2)!/ (n-1)! = 210
A),B),C)
It's like finding the number of ways to arrange 4 1's and 7-0's.
The number of ways is given by (m+n)!/(m!n!)
Experiment first with a grid of 2x2, and then a 2x3.
2.
(n-r)!/(n-r+1)!
=(1.2.3...n-r)/(1.2.3...(n-r).(n-r+1))
=1/(n-r+1)
3.
similarly,
(n+2)!/(n-1)! = 210
(1.2.3...(n+2))/(1.2.3...(n-1)) = 210
n.(n+1).(n+2) = 210
Start with cube root of 210 = 5.9...
so try 5.6.7=210 OK.
So n=5,n+1=6,n+2=7
thank even though im not sure if u woule see this
You're welcome!
1) To find the number of grid paths from the origin in a coordinate graph to a specific point, we can use the concept of combinations. Since we can only move either right (R) or up (U) in a grid, the number of paths can be calculated using the formula for combinations.
a) To find the number of paths to point (4,7), we need to calculate the combination of 4 right moves (R) out of a total of 11 moves (7 up (U) + 4 right (R)). The formula for combinations is given as C(n, r) = n! / (r!(n-r)!), where n is the total number of moves and r is the number of right moves. In this case, we have C(11, 4) = 11! / (4!(11-4)!).
b) To find the number of paths to point (3,7), we need to calculate the combination of 3 right moves (R) out of a total of 10 moves (7 up (U) + 3 right (R)). Using the combination formula, we get C(10, 3) = 10! / (3!(10-3)!).
c) To find the number of paths to point (4,6), we need to calculate the combination of 4 right moves (R) out of a total of 10 moves (6 up (U) + 4 right (R)). Again, using the combination formula, we get C(10, 4) = 10! / (4!(10-4)!).
2) To simplify the expression (n-r)! / (n-r+1)!, we need to simplify the factorial terms. Recall that (n-r)! = (n-r) * (n-r-1) * ... * 1, and (n-r+1)! = (n-r+1) * (n-r) * ... * 1.
Canceling out the common terms, we get (n-r)! / (n-r+1)! = (n-r) * (n-r-1) * ... * 1 / [(n-r+1) * (n-r) * ... * 1]. We can notice that all terms from (n-r) down to 1 cancel out, leaving us with 1 / (n-r+1).
Therefore, the simplified expression is 1 / (n-r+1).
3) To solve the equation (n+2)! / (n-1)! = 210, we can simplify the factorial terms as explained in the previous question.
Expanding the factorial terms, (n+2)! = (n+2) * (n+1) * n! and (n-1)! = (n-1) * (n-2) * ... 1.
Substituting these values into the equation, we get [(n+2) * (n+1) * n!] / [(n-1) * (n-2) * ... 1] = 210.
Simplifying further, we can cross-cancel common terms, giving us (n+2) * (n+1) = 210.
Expanding the product, we get n^2 + 3n + 2 = 210.
Rearranging the equation and simplifying, we have n^2 + 3n - 208 = 0.
Now, we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. Once we find the values of n, we can verify if they satisfy the original equation.
Note: The steps mentioned above are general techniques to arrive at the solution. The specific method may vary depending on the equation and its complexity.