2013 MMC Elimination for 4th Year Combinatorics Questions Solution

Below are the questions in Combinatorics taken from 2013 Metrobank-MTAP-Dep-Ed Math Challenge elimination round. This post is created to be guide dedicated students preparing for the nationwide Math Challenge. All 50 problems can be found here.

Question 8:

How many 5-digit even numbers can be formed out of the digits 2,4,5,8,and 9 (without repetition)?

Solution:

Consider the 5-digit number as blanks:  _ _ _ _ _

The last digit must be even. Since we have three even numbers (2,4,8). The last digit can be filled in 3 ways. Since repetition is not allowed, we can fill out the rest blanks in ways,3 ways,2 ways, and 1 way.

1 x 2 x 3 x 4 x 3 = 72 ways.

Question 24:

If a fair coin is tossed five times, find the probability that exactly 3 tosses show heads.

Solution:

This topic is under probability of repeated trials. Mathematicians often call it “Binomial” because this is an application of binomial theorem in probability.

Let H be the Head outcome of a coin

Let T be the Tail outcome of a coin

One way to arrange the given condition is the sequence: H1|H2|H3|T1|T2

But the following sequence also meets the condition: H2|H1|H3|T1|T2 or H2|H3|T1|T2|H1.

Counting all possible outcomes is the same technique used in counting the number of permutations in the word BOOK.

All possible outcomes can be calculated in  $\displaystyle\frac{5!}{3!2!}=10$ ways. Each coin has 2 ways to land. Thus 5 coins have 2x2x2x2x2=32 ways to land. Hence, the desired probability is  $\displaystyle\frac{10}{32}=\boxed{\displaystyle\frac{5}{16}}$

Question 32:

A bookshelf has 8 history books and 10 cooking books. You will select 10 books ( 2 History books and 8 cooking books) to bring on a trip. How many choices are possible?

Solutions:

In the language of combinations, we can select 2 history books out of 8 in 8C2 = 28 ways. Likewise, we can select 8 cooking books out of 10 cooking books in 10C8 = 45 ways. By counting principle, there are 28×45=1260 possible choices.

Question 41:

The dial on a combination lock contains three wheels, each of which is labeled with a digit from 0 to 9. How many possible combinations does the lock have if digits may not be repeated?

Solution:

Let A,B,C stands for digit from 0-9 as a lock combination. A can filled in 10 ways (0-9). B can be filled in 9 ways, C can be filled in 8 ways. By counting principle, there are 10x9x8=720 to possible combinations of lock.

Question 46:

If four-number codes are form randomly from the digits 0 to 9, what is the probability that the two middle digits are the same?

Solution:

Consider the sequence A-B-B-C. There is no given condition that the digits can’t be repeated so we assume that we can repeat each digit. Each letter can represent 10 numbers (0-9). But 2 B’s is counted as one. Thus the total possible ways is 10x10x10=1,000.

The total number of possible combinations is 10x10x10x10=10,000. The probability that the two middle digits are the same is 1,000/10,000=1/10.