Some types of word problems are easily solved using Combinations:
Suppose that you are going to choose a small group of 3 items
from a larger group of 7 items.
You could list all the possible groups, if need be.
But, you can count the number of the possible groups
by just using COMBINATIONS.
7 choose 3 has _35__ possible groups
7! / (3! times 4!) = 35
(7*6*5*4*3*2*1*)
(3*2*1)(4*3*2*1)
Reduce by dividing top
and bottom by (4!)
= (7*6*5)/(3*2*1)
OR
LOOKING at PASCAL's TRIANGLESee the row with 1, 7, 21, 35, 35, 21, 7, 1?
7 choose NONE has _1_ possible group
7 choose 1 has _7_ possible groups
7 choose 2 has _21_ possible groups
7 choose 3 has _35_ possible groups
7 choose 4 has _35_ possible groups
7 choose 5 has _21_ possible groups
7 choose 6 has _7_ possible groups
7 choose 7 has _1_ possible groups
HOW ABOUT using the TI-83?
7 choose NONE has _1_ possible group
7 choose 1 has _7_ possible groups
7 choose 2 has _21_ possible groups
7 choose 3 has _35_ possible groups
7 choose 4 has _35_ possible groups
7 choose 5 has _21_ possible groups
7 choose 6 has _7_ possible groups
7 choose 7 has _1_ possible groups
**************************************
This forms Sierpiński triangle
Learn more at:
You remember the area code and first 3 digits of a phone number.
How many ways can the last four digits appear?
How many ways, if you know the four different digits but not the order?
you know they are 0034 but not the order?
2224 but not the order?
How many zip codes are possible?
what if 0 can not be the first?
what if all are different digits?
All different and the first not a 0?
Suppose we wish to arrange n = 5 people {a, b, c, d, e}, standing side by side, for
a portrait. How many such distinct portraits (“permutations”) are possible?
5! = 120.
Permutations of {a, b, c, d, e}
These are the 5! = 120 ways of arranging 5 objects,
in such a way that all the different orders count as being distinct.
a b c d e b a c d e c a b d e d a b c e e a b c d
a b c e d b a c e d c a b e d d a b e c e a b d c
a b d c e b a d c e c a d b e d a c b e e a c b d
a b d e c b a d e c c a d e b d a c e b e a c d b
a b e c d b a e c d c a e b d d a e b c e a d b c
a b e d c b a e d c c a e d b d a e c b e a d c b
a c b d e b c a d e c b a d e d b a c e e b a c d
a c b e d b c a e d c b a e d d b a e c e b a d c
a c d b e b c d a e c b d a e d b c a e e b c a d
a c d e b b c d e a c b d e a d b c e a e b c d a
a c e b d b c e a d c b e a d d b e a c e b d a c
a c e d b b c e d a c b e d a d b e c a e b d c a
a d b c e b d a c e c d a b e d c a b e e c a b d
a d b e c b d a e c c d a e b d c a e b e c a d b
a d c b e b d c a e c d b a e d c b a e e c b a d
a d c e b b d c e a c d b e a d c b e a e c b d a
a d e b c b d e a c c d e a d d c e a b e c d a b
a d e c b b d e c a c d e d a d c e b a e c d b a
a e b c d b e a c d c e a b d d e a b c e d a b c
a e b d c b e a d c c e a d b d e a c b e d a c b
a e c b d b e c a d c e b a d d e b a c e d b a c
a e c d b b e c d a c e b d a d e b c a e d b c a
a e d b c b e d a c c e d a b d e c a b e d c a b
a e d c b b e d c a c e d b a d e c b a e d c b a
Now suppose we start with the same n = 5 people {a, b, c, d, e}, but we wish to
make portraits of only k = 3 of them at a time. How many such distinct portraits are possible?
5 × 4 × 3 =
5!/
2! .
Finally suppose that instead of portraits (“permutations”), we wish to form
committees (“combinations”) of k = 3 people from the original n = 5. How many such distinct committees are possible?
5!
2! , divided by 3!, i.e., 60 ÷ 6 = 10. “n choose k”.
Suppose you toss a coin n = 5 times in a row. How many ways can you end up with k = 3 heads?
are
5
3
? ? ? ? ? ?
= 10 ways of obtaining k = 3 heads in n = 5 independent successive tosses.
A license plate begins with three letters. If the possible letters are A, B, C, D and E, how many different permutations of these letters can be made if no letter is used more than once?
The problem involves 5 things (A, B, C, D, E) taken 3 at a time.
There are 60 different permutations for the license plate.
In how many ways can a president, a treasurer and a secretary be chosen from among 7 candidates?
Solution:
Using reasoning:
For the first position, there are 7 possible choices. After that candidate is chosen, there are 6 possible choices. Finally, there are 5 possible choices.
7 × 6 × 5 = 210
A zip code contains 5 digits. How many different zip codes can be made with the digits 0–9 if no digit is used more than once and the first digit is not 0?
Solution:
Using reasoning:
For the first position, there are 9 possible choices (since 0 is not allowed). After that number is chosen, there are 9 possible choices (since 0 is now allowed). Then, there are 8 possible choices, 7 possible choices and 6 possible choices.
9 × 9 × 8 × 7 × 6 = 27,216
Using permutation formula:
We can’t include the first digit in the formula because 0 is not allowed.
For the first position, there are 9 possible choices (since 0 is not allowed). For the next 4 positions, we are selecting from 9 digits.
There are two cases of circular-permutations:-
(a) If clockwise and anti clock-wise orders are different, then total number of circular-permutations is given by (n-1)!
(b) If clock-wise and anti-clock-wise orders are taken as not different, then total number of circular-permutations is given by (n-1)!/2!
No comments:
Post a Comment