SOLUTION: The object is to start with the letter A on top and to move down the diagram to the C at the bottom. From any given letter, move only to one of the letters directly below it on the
Algebra ->
Probability-and-statistics
-> SOLUTION: The object is to start with the letter A on top and to move down the diagram to the C at the bottom. From any given letter, move only to one of the letters directly below it on the
Log On
Question 1026650: The object is to start with the letter A on top and to move down the diagram to the C at the bottom. From any given letter, move only to one of the letters directly below it on the left or right. If these rules are followed, how many different paths spell ALGEBRAIC?
The first row is a single A.
The second row contains two L's.
The third row: 3 G's.
The fourth row: 4 E's.
The fifth row: 5 B's.
The sixth row: 4 R's.
The seventh row: 3 A's.
The eight row: 2 I's.
The ninth row: 1 C.
The shape produces a diamond. Answer by Edwin McCravy(20054) (Show Source):
A
L L
G G G
E E E E
B B B B B
R R R R
A A A
I I
C
Look at a sample path:
A
/
L L
\
G G G
\
E E E E
\
B B B B B
/
R R R R
/
A A A
\
I I
/
C
That particular path could be represented by LRRRLLRL.
Or you could stay along one of the edges:
A
/
L L
/
G G G
/
E E E E
/
B B B B B
\
R R R R
\
A A A
\
I I
\
C
That path could be represented by LLLLRRRR
Every path could be represented by a distinguishable
permutation of LLLLRRRR. And every distinguishable
permutation of LLLLRRRR would represent a different
path.
You can look at that either of two ways:
The number of distinguishable permutations of LLLLRRRR
and calculate it this way:
So the answer is possible paths.
Or we could say there are 8 positions in each distinguishable
permutation of LLLLRRRR, and we choose 4 or those positions to
place the L's in in 8C4 ways and the other 4 would be filled
with R's in only 1 way. So the answer in that case would be
simply the combinations of 8 things taken 4 at a time.
8C4 = 70.
Edwin
