6.1 Q-6 | Permutation, Combination And Probability | Mathematics 11

Exercise 6.1 — Question 6

This question involves evaluating or simplifying expressions using the combination formula.

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

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

Evaluate $(3 10)$ :

$(3 10) = \frac{10 !}{3 ! \cdot 7 !} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120$

Verify the symmetry property: $(7 10) = (3 10) = 120$ ✓

If $(2 n) = 15$ , find $n$ :

$\frac{n !}{2 ! ( n - 2 )!} = 15 ⟹ \frac{n ( n - 1 )}{2} = 15 ⟹ n (n - 1) = 30 ⟹ n = 6$

This question involves evaluating or simplifying expressions using the combination formula.

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

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

Evaluate $(3 10)$ :

$(3 10) = \frac{10 !}{3 ! \cdot 7 !} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = \frac{720}{6} = 120$

Verify the symmetry property: $(7 10) = (3 10) = 120$ ✓

If $(2 n) = 15$ , find $n$ :

$\frac{n !}{2 ! ( n - 2 )!} = 15 ⟹ \frac{n ( n - 1 )}{2} = 15 ⟹ n (n - 1) = 30 ⟹ n = 6$