In combinatorial mathematics, a k-combination of a finite set S is a subset of k different numbers of S. Specifying a subset does not arrange them in a particular order; by contrast, producing the k different numbers in a specific order defines a sequence without repetition, also called k-permutation (but which is not a permutation of S in the usual sense of that term. SOURCE: WIKIPEDIA
Example problems of different number combinations:
Different number combinations example 1:
How many lines can you draw using THREE non collinear (not in a single line) points X, Y and Z on a plane?
Solution:
You need two points to draw a line. The order is not important. Line XY is the same as line YX. The problem is to select TWO points out of THREE to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.
XY, XZ
YX, YZ
ZX, ZY
There is a problem: line XY is the same as line YX, same for lines XZ and ZX and YZ and ZY.
The lines are: XY, YZ and XZ; three lines only.
So in fact we can draw THREE lines and not SIX and that's because in this problem the order of the points X, Y and Z is not important.
This is a combination problem: combining TWO items out of THREE and is written the general form as follows:
n C r = n! / [ (n - r)! r! ]
Special case:
n C 0 = n C n = 1
The number of combinations (nCr) is equal to the number of permutations divided by r! to removes those counted more than once because the order is not mainly use.
Different number combinations example 2:
Calculate 4C3
Solution:
We can find the combination value by using the following formula:
n C r = n! / [(n - r)! r!]
Substitute the value of n and r into the above formula, then we get
4C3= 4! / [(4-3)! 3!]
=4! / [1!*3!]
=24/[1*6]
=24/6
=4
Answer: 4
Different number combinations example 3:
Calculate 6C6
Solution:
We can find the combination value by using the following formula:
n C r = n! / [(n - r)! r!]
Substitute the value of n and r into the above formula, then we get
6C6= 6! / [(6-6)! 6!]
=6! / [0!*6!]
=6! / [1*(6!)]
=6! / 6!
= 1
Answer: 1
Having problem with Probability Sets Read my upcoming post, i will try to help you.
Practice problems of different number combinations:
1) In how many ways can you select a committee of 2 students out of 5 students?
Answer: 20 (5C2)
2) How many triangles can you make using 4 non collinear points on a plane?
Answer: 24 (4C3)
Example problems of different number combinations:
Different number combinations example 1:
How many lines can you draw using THREE non collinear (not in a single line) points X, Y and Z on a plane?
Solution:
You need two points to draw a line. The order is not important. Line XY is the same as line YX. The problem is to select TWO points out of THREE to draw different lines. If we proceed as we did with permutations, we get the following pairs of points to draw lines.
XY, XZ
YX, YZ
ZX, ZY
There is a problem: line XY is the same as line YX, same for lines XZ and ZX and YZ and ZY.
The lines are: XY, YZ and XZ; three lines only.
So in fact we can draw THREE lines and not SIX and that's because in this problem the order of the points X, Y and Z is not important.
This is a combination problem: combining TWO items out of THREE and is written the general form as follows:
n C r = n! / [ (n - r)! r! ]
Special case:
n C 0 = n C n = 1
The number of combinations (nCr) is equal to the number of permutations divided by r! to removes those counted more than once because the order is not mainly use.
Different number combinations example 2:
Calculate 4C3
Solution:
We can find the combination value by using the following formula:
n C r = n! / [(n - r)! r!]
Substitute the value of n and r into the above formula, then we get
4C3= 4! / [(4-3)! 3!]
=4! / [1!*3!]
=24/[1*6]
=24/6
=4
Answer: 4
Different number combinations example 3:
Calculate 6C6
Solution:
We can find the combination value by using the following formula:
n C r = n! / [(n - r)! r!]
Substitute the value of n and r into the above formula, then we get
6C6= 6! / [(6-6)! 6!]
=6! / [0!*6!]
=6! / [1*(6!)]
=6! / 6!
= 1
Answer: 1
Having problem with Probability Sets Read my upcoming post, i will try to help you.
Practice problems of different number combinations:
1) In how many ways can you select a committee of 2 students out of 5 students?
Answer: 20 (5C2)
2) How many triangles can you make using 4 non collinear points on a plane?
Answer: 24 (4C3)
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