IB Math HL – Probability & Combinatorics

Questions and Answers

1. Permutations Combinations – IB Mathematics HL

How can we find the number of possible choices for choosing a sub-committee of five persons consisting of the chairperson, secretary and three ordinary membersfrom a committee of 40 people?

Solution

Permutations, Combinations – IB Maths HL

The chairperson can be chosen by $^{40}\mathrm{C}_1$ ways

The secretary can be chosen by $^{39}\mathrm{C}_1$ ways

And the three ordinary members can be chosen by $^{38}\mathrm{C}_3$ ways

So, the members of this sub-committee can be chosen by

$^{40}\mathrm{C}_1 \cdot ^{39}\mathrm{C}_1 \cdot ^{38}\mathrm{C}_3=$

$=40 \cdot 39\cdot8436=13,160,160$ ways.

2. Permutations Combinations – IB Math HL

In a class of 12 students, 4 are male and 8 are female. How can we find the number of committees of 5 members can be formed containing 2 males and 3 females?

Solution

Permutations, Combinations – IB Maths HL

The males can be chosen by $^4\mathrm{C}_2$ ways

The females, can be chosen by $^8\mathrm{C}_3$ ways

So, the members of this committee can be chosen by

$^4\mathrm{C}_2 \cdot ^8\mathrm{C}_3=$

$=6 \cdot 56=336$ ways.

3. Permutations Combinations Probability- IB Mathematics HL

How can we find the number of different arrangements of the letters of the word “LONDON”? and how can we calculate the probability of the event that the middle two letters both are N’s?

Solution

Permutations, Combinations, Probability – IB Maths HL

The number of different arrangements of the six letters from which there are two pairs of same letters is given by the following formula

$\frac{6!}{2!2!}=\frac{720}{4}=180$

Since the two N’s are located in the middle of the word there are $\frac{4!}{2!}=\frac{24}{2}=12$ ways to arrange the remaining 4 letters from which there are two O,s.