6.1 Q-4 | Permutation, Combination And Probability | Mathematics 11

Exercise 6.1 — Question 4: Combinations

The number of ways to choose $r$ objects from $n$ distinct objects without regard to order is given by:

$^{n} C_{r} = (r n) = \frac{n !}{r ! ( n - r )!}$

where $n \geq r \geq 0$ .

Example 1: Evaluate $^{8} C_{3}$ .

$^{8} C_{3} = \frac{8 !}{3 ! \cdot 5 !} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = \frac{336}{6} = 56$

Example 2: Find $n$ if $^{n} C_{2} = 15$ .

$\frac{n !}{2 ! ( n - 2 )!} = 15 ⟹ \frac{n ( n - 1 )}{2} = 15 ⟹ n (n - 1) = 30$

Since $6 \times 5 = 30$ , we get $n = 6$ .

Example 3: Use the symmetry property to simplify $^{12} C_{10}$ .

$^{12} C_{10} =^{12} C_{12 - 10} =^{12} C_{2} = \frac{12 \times 11}{2} = 66$

Example 4: In how many ways can a committee of 3 be selected from 7 people?

$^{7} C_{3} = \frac{7 !}{3 ! \cdot 4 !} = \frac{7 \times 6 \times 5}{6} = 35 ways$

Permutation $^{n} P_{r}$ : order matters — arranging $r$ from $n$ .
Combination $^{n} C_{r}$ : order does not matter — selecting $r$ from $n$ .

$^{n} C_{r} = \frac{^{n} P _{r}}{r !}$