Permutations and Combinations Miscellaneous Excercise
Question:
It is required to seat 5 men and 4 women in a row so that the women occupy the even places. How many such arrangements are possible?
Answer:
Step 1: There are 4 women and 5 men.
Step 2: The women must occupy the even places, which means the 1st, 3rd, 5th and 7th places.
Step 3: The men must occupy the odd places, which means the 2nd, 4th, 6th and 8th places.
Step 4: The number of possible arrangements is the number of ways to arrange 5 men in 4 odd places, which is equal to 5!/(54)! = 5!.
Step 5: Therefore, the number of possible arrangements is 5! = 120.
Question:
An examination paper consists of 12 questions divided into two parts A & B. Part A, containing 7 questions & part B contains 5 questions. A candidates is required to attempt 8 questions selecting at least 3 from each part. In how many ways can the candidates select the questions?
Answer:
Answer:
 Total number of questions = 12
 Part A has 7 questions and Part B has 5 questions
 A candidate is required to attempt 8 questions, selecting at least 3 from each part
 Number of ways to select 3 questions from Part A = 7C3 = 35
 Number of ways to select 5 questions from Part B = 5C5 = 1
 Therefore, the total number of ways to select questions = 35 x 1 = 35
Question:
From a class of 25 students, 10 are to be chosen for an excursion party. There are 3 students who decide that their all of them will join or none of them will join. In how many ways can the excursion party be chosen?
Answer:
Answer:
 There are 25 students in the class.
 10 students need to be chosen for the excursion party.
 There are 3 students who decide that their all of them will join or none of them will join.
Therefore, there are two possible scenarios:
Scenario 1: The 3 students join the excursion party: In this case, the excursion party can be chosen in $\binom{22}{7}$ ways.
Scenario 2: The 3 students do not join the excursion party: In this case, the excursion party can be chosen in $\binom{22}{10}$ ways.
Hence, the total number of ways in which the excursion party can be chosen is $\binom{22}{7} + \binom{22}{10}$ ways.
Question:
In how many ways can be letters of word ASSASSINATION be arranged so that all the S’s are together?
Answer:
Answer:
Step 1: Arrange the letters of the word ASSASSINATION in alphabetical order: AACIINNOSSST
Step 2: Group all the S’s together: SSSAAACIINNO
Step 3: Calculate the total number of ways the letters can be arranged: 8!/(3!2!) = 336
Question:
How many words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together?
Answer:

First, count the number of letters in the word EQUATION. There are 8 letters.

Determine the number of vowels and consonants in the word EQUATION. There are 4 vowels (E, U, A, O) and 4 consonants (Q, T, I, N).

Calculate the number of words that can be formed using all 8 letters at a time. This can be done by multiplying the number of vowels and consonants together: 4 x 4 = 16.

Since the vowels and consonants must occur together, the number of words that can be formed is half of the total: 16/2 = 8.
Therefore, 8 words, with or without meaning, can be formed using all the letters of the word EQUATION at a time so that the vowels and consonants occur together.
Question:
If the different permutations of all the letter of the words EXAMINATION are listed as in a dictionary, how many words are there in this list before the first word starting with E?
Answer:
Step 1: Count the total number of letters in the word EXAMINATION. Answer: 10
Step 2: Determine the total number of permutations of the 10 letters in the word EXAMINATION. Answer: 10! = 3,628,800
Step 3: Calculate the number of words in the list that start with a letter other than E. Answer: There are (26  1) x 10! = 3,606,720 words in the list that start with a letter other than E.
Step 4: Subtract the number of words in the list that start with a letter other than E from the total number of permutations of the 10 letters in the word EXAMINATION. Answer: 3,628,800  3,606,720 = 22,080 words in the list before the first word starting with E.
Question:
How Many words, with or without meaning, each of 2 vowels and 3 consonants can be formed from the letters of the word DAUGHTER?
Answer:
Answer: Step 1: Calculate the total number of letters in the word DAUGHTER. Answer: 8
Step 2: Calculate the number of combinations possible with 2 vowels and 3 consonants. Answer: The number of combinations possible is 8C3 = 56.
Step 3: Calculate the total number of words that can be formed. Answer: The total number of words that can be formed is 56.
Question:
The English alphabet has 5 vowels and 21 consonants. How many words with two different vowels and 2 different consonants can be formed from the alphabet?
Answer:
Step 1: Count the total number of letters available in the alphabet.
5 vowels + 21 consonants = 26 letters
Step 2: Calculate the number of combinations of two different vowels and two different consonants.
Number of combinations = 5 × 4 × 21 × 20 = 16800
Step 3: Calculate the total number of words that can be formed from the combinations.
Number of words = 16800 × 2! = 33600
Question:
A committee of 7 has to be formed from 9 boys and 4 girls. In how many ways can this be done when the committee consists of: (i) exactly 3 girls? (ii) at least 3 girls? (iii) at most 3 girls?
Answer:
(i) Exactly 3 girls:
There are 4 girls and 9 boys, so the total number of people to choose from is 13.
We need to choose 3 girls from the 4 available girls. This can be done in 4C3 ways, which is equal to 4.
We now need to choose 4 boys from the 9 available boys. This can be done in 9C4 ways, which is equal to 126.
Therefore, the total number of ways to form a committee of 7 with exactly 3 girls is 4 x 126 = 504.
(ii) At least 3 girls:
We need to choose 3 girls from the 4 available girls. This can be done in 4C3 ways, which is equal to 4.
We now need to choose 4 boys from the 9 available boys. This can be done in 9C4 ways, which is equal to 126.
We also need to consider the case when there are 4 girls in the committee. This can be done in 4C4 ways, which is equal to 1.
We now need to choose 3 boys from the 9 available boys. This can be done in 9C3 ways, which is equal to 84.
Therefore, the total number of ways to form a committee of 7 with at least 3 girls is 4 x 126 + 1 x 84 = 616.
(iii) At most 3 girls:
We need to consider the cases when there are 0, 1, 2 and 3 girls in the committee.
Case 1: 0 girls
We need to choose 7 boys from the 9 available boys. This can be done in 9C7 ways, which is equal to 36.
Case 2: 1 girl
We need to choose 1 girl from the 4 available girls. This can be done in 4C1 ways, which is equal to 4.
We now need to choose 6 boys from the 9 available boys. This can be done in 9C6 ways, which is equal to 84.
Case 3: 2 girls
We need to choose 2 girls from the 4 available girls. This can be done in 4C2 ways, which is equal to 6.
We now need to choose 5 boys from the 9 available boys. This can be done in 9C5 ways, which is equal to 126.
Case 4: 3 girls
We need to choose 3 girls from the 4 available girls. This can be done in 4C3 ways, which is equal to 4.
We now need to choose 4 boys from the 9 available boys. This can be done in 9C4 ways, which is equal to 126.
Therefore, the total number of ways to form a committee of 7 with at most 3 girls is 36 + 4 x 84 + 6 x 126 + 4 x 126 = 864.
Question:
How many 6−digit numbers can be formed from the digits 0,1,3,5,7 and 9 which are divisible by 10 and no digit is repeated?
Answer:
Answer:

There are six digits 0,1,3,5,7 and 9.

No digit should be repeated, so the 6digit numbers should be unique.

The numbers should be divisible by 10.

Therefore, the numbers should end with 0.

So, the last digit is already fixed as 0.

Now, we have to arrange the remaining 5 digits in such a way that the number is divisible by 10.

The sum of the digits should be divisible by 10.

The possible combinations are 01357, 01379, 03157, 03179, 05137, 05139, 07135, 07139, 09135, 09137.

Therefore, there are 10 6digit numbers that can be formed from the digits 0,1,3,5,7 and 9 which are divisible by 10 and no digit is repeated.
Question:
Determine the number of 5card combination out of a deck of 52 cards if each selection of 5 cards has exactly one king.
Answer:
Answer: Step 1: There are 4 kings in a deck of 52 cards.
Step 2: The number of 5card combinations that include exactly one king is equal to the number of ways to select 4 cards from the remaining 48 cards multiplied by the number of ways to select 1 king from the 4 kings.
Step 3: The number of ways to select 4 cards from the remaining 48 cards is equal to 48C4.
Step 4: The number of ways to select 1 king from the 4 kings is equal to 4C1.
Step 5: The number of 5card combinations that include exactly one king is equal to 48C4 x 4C1 = 10,096.