Combinatorics is a branch of mathematics that deals with counting, selecting, and arranging objects. It provides methods to determine how many different ways items can be chosen or arranged without listing every possibility. It includes concepts such as permutations and combinations.
Question 1: How many words with 2 different vowels and 2 different consonants can be formed from the alphabet?
Solution:
To solve this problem, we have to know that the alphabet contains 26 letters including 5 vowels and 21 consonants.
Step 1: Ways of selecting 2 different vowels out of 5 vowels
C(5, 2) = 5! / (2! × (5 – 2)!)
= 5! / (2! × 3!) = 10
Step 2: Ways of selecting 2 different consonants out of 21 consonants
C(21, 3) = 21! / (2! × (21 – 2)!)
= 21! / (3! × 19!) = 210
Hence, combinations of 2 different vowels and 3 different consonants is 10 × 210 = 2100
Step 3: Ways arranging or forming 2 vowels and 2 consonants out of 4 letters
P(4, 4) = 4! / (4 – 4)!
= 4! / 0! = 4! = 24
Therefore, the total number of ways is 2100 × 24 = 54000
Answer: 54000 words
Question 2: How many distinct ways are there to arrange 10 people in a row if exactly 3 people must be together in one block?
Solution:
To solve this problem, we need to treat the block of 3 people who must be together as a single unit.
The 3 people within this block can be arranged among themselves in 3! ways. [3! = 6]
Since the 3 people are treated as one unit, we now have 8 units to arrange (the block plus the other 7 individual people).
The 8 units can be arranged in (8!) ways. [8! = 40320]
To find the total number of distinct ways to arrange the 10 people with the block treated as a single unit, multiply the number of ways to arrange the units by the number of ways to arrange the people within the block.
[8! × 3! = 40320 × 6 = 241920]
Therefore, there are (241,920) distinct ways to arrange 10 people in a row if exactly 3 of them must be together in one block.
Question 3: How many 5-digit numbers can be formed such that the digit 3 appears exactly twice?
Solution:
To determine how many 5-digit numbers can be formed such that the digit 3 appears exactly twice, we can break down the problem as follows:
Choose the positions for the two 3's:
We need to choose 2 positions out of 5 for the digit 3. The number of ways to choose 2 positions out of 5 is given by the combination formula
5C2 = 5!/2!(5-2)! = 10
The remaining 3 positions can be filled with any digits from 0 to 9 except for 3. Since we are forming 5-digit numbers, the first digit cannot be 0.
The first digit (one of the remaining three positions) can be any digit from 1 to 9 except 3. That gives us 8 choices for the first digit.
The other two positions can be any digit from 0 to 9 except 3. That gives us 9 choices for each of these two positions.
For the first position (not 0 and not 3), we have 8 choices.
For the second position (can be 0 but not 3), we have 9 choices.
For the third position (can be 0 but not 3), we have 9 choices.
Therefore, the number of ways to fill these remaining three positions is: 8 × 9 × 9 = 648
We now combine the number of ways to choose the positions for the 3's and the number of ways to fill the remaining positions.
10 × 648 = 6480
Therefore, there are (6,480) different 5-digit numbers where the digit 3 appears exactly twice.
Question 4: How many ways can you arrange the letters in the word "MATHEMATICS" such that the two 'M's are next to each other?
Solution:
Treat two 'M's as a single unit:
This reduces the problem to arranging the units: "MM", A, T, H, E, A, T, I, C, S (10 units).
Count permutations of these units:
Frequencies:
A: 2
T: 2
Other letters appear once.
Use the formula: 10!/2!×2!
Calculate: 3628800/4=907200
Question 5: In how many ways can you choose 4 books from a shelf of 12 books?
Solution:
4 Books from the shelf of 12 books can be choose in 4C12 number of ways.
Use the combination formula
12C4 = 12!//4!(12−4)!
Calculate = (12×11×10×9)/(4×3×2×1) = 495
There are 495 ways to choose 4 books from 12.
Practice Problems
Problem 1: How many distinct ways can you arrange 7 people around a circular table?
Problem 2: How many ways can you distribute 5 identical candies into 3 distinct boxes?
Problem 3: In how many ways can you arrange 4 items so that none of them is in its original position?
Problem 4: How many ways can you arrange 6 people around a circular table?
Problem 5: How many ways can you distribute 5 identical candies to 3 children?
