Combination

 4.2 Combination

Difference between Permutation and Combination
 Permutation Combination Process of arranging objects where order and sequence are taken into consideration. Process of selection without considering the order and sequence of the objects.

 Number of Combinations of $$r$$ Objects Chosen from $$n$$ Different Objects at a Time $${}^nC_r=\dfrac{{}^nP_r}{r!}=\dfrac{n!}{r!(n-r)!}$$

Example
 Question
 (a) Find the number of triangles that can be formed from the vertices of a hexagon. (b) Encik Samad wants to choose three types of batik motifs from four organic motifs and five geometrical motifs. Find the number of ways to choose at least one organic motif and one geometrical motif.
 Solution

(a)

Hexagon has six vertices.

To form a triangle, any three vertices are required.

So, the number of ways is

\begin{aligned} {}^nC_{r} &= \dfrac{6!}{3!(6-3)!} \\\\ &=\dfrac{6!}{3!3!}\\\\ &=20. \end{aligned}

(b)

Number of ways to choose two organic motifs and one geometric motif,  $${}^4C_{2} \times{}^5C_{1}$$.

Number of ways to choose one organic motif and two geometric motifs,  $${}^4C_{1} \times {}^5C_{2}$$.

So, the number of ways

$$({}^4C_{2} \times {}^5C_{1}) + ({}^4C_{1} \times {}^5C_{2}) = 70$$.

Combination

 4.2 Combination

Difference between Permutation and Combination
 Permutation Combination Process of arranging objects where order and sequence are taken into consideration. Process of selection without considering the order and sequence of the objects.

 Number of Combinations of $$r$$ Objects Chosen from $$n$$ Different Objects at a Time $${}^nC_r=\dfrac{{}^nP_r}{r!}=\dfrac{n!}{r!(n-r)!}$$

Example
 Question
 (a) Find the number of triangles that can be formed from the vertices of a hexagon. (b) Encik Samad wants to choose three types of batik motifs from four organic motifs and five geometrical motifs. Find the number of ways to choose at least one organic motif and one geometrical motif.
 Solution

(a)

Hexagon has six vertices.

To form a triangle, any three vertices are required.

So, the number of ways is

\begin{aligned} {}^nC_{r} &= \dfrac{6!}{3!(6-3)!} \\\\ &=\dfrac{6!}{3!3!}\\\\ &=20. \end{aligned}

(b)

Number of ways to choose two organic motifs and one geometric motif,  $${}^4C_{2} \times{}^5C_{1}$$.

Number of ways to choose one organic motif and two geometric motifs,  $${}^4C_{1} \times {}^5C_{2}$$.

So, the number of ways

$$({}^4C_{2} \times {}^5C_{1}) + ({}^4C_{1} \times {}^5C_{2}) = 70$$.