### Permutations and Combinations Exercise 03

## Question:

In how many ways can the letters of the word PERMUTATIONS be arranged if the.i) Words start with P and end with S.ii) vowels are all togetheriii) there are always 4 letters between P and S

## Answer:

i) There are 8!/(2!2!) = 2520 ways to arrange the letters of the word PERMUTATIONS if it starts with P and ends with S.

ii) There are 8!/(2!2!2!) = 5040 ways to arrange the letters of the word PERMUTATIONS if the vowels are all together.

iii) There are 8!/(4!2!2!) = 1260 ways to arrange the letters of the word PERMUTATIONS if there are always 4 letters between P and S.

## Question:

How many 4-digit numbers are there with no digit repeated?

## Answer:

Step 1: A 4-digit number is composed of 4 digits, each ranging from 0-9.

Step 2: There are 10 possible choices for the first digit, 9 choices for the second digit, 8 choices for the third digit, and 7 choices for the fourth digit.

Step 3: To find the total number of 4-digit numbers with no digit repeated, we can multiply the number of choices for each digit: 10 x 9 x 8 x 7 = 5040.

Therefore, there are 5040 4-digit numbers with no digit repeated.

## Question:

How many 3-digit even numbers can be made using the digits 1,2,3,4,6,7 if no digit is repeated?

## Answer:

Answer:

Step 1: The total number of 3-digit even numbers that can be made using the digits 1,2,3,4,6,7 is 6C3 = 20.

Step 2: To make sure that no digit is repeated, we will need to use the digits 2,4,6.

Step 3: The total number of 3-digit even numbers that can be made using the digits 2,4,6 is 3C3 = 1.

Therefore, the answer is 1.

## Question:

How many words, with or without meaning can be made from the letters of the word MONDAY, assuming that no letter is repeated, if. (i) 4 letters are used at a time.(ii) all letters are used at a time(iii) all letters are used but first letter is a vowel?

## Answer:

(i) 4 letters are used at a time:

There are 6 possible combinations of 4 letters from the word MONDAY:

MOND, MONA, ONDA, ONDY, NDAY, and NDA.

Therefore, 6 words can be made from the letters of MONDAY if 4 letters are used at a time.

(ii) All letters are used at a time:

There is only 1 possible combination of all the letters from the word MONDAY: MONDAY.

Therefore, 1 word can be made from the letters of MONDAY if all letters are used at a time.

(iii) All letters are used but first letter is a vowel:

There are 5 possible combinations of all the letters from the word MONDAY, but with the first letter being a vowel:

ONDAY, ONDYA, ONDAY, ODNAY, and ODNYA.

Therefore, 5 words can be made from the letters of MONDAY if all letters are used but the first letter is a vowel.

## Question:

There are always 4 letters between P and S?

## Answer:

Step 1: Determine the letters between P and S. Answer: Q, R, T, U

## Question:

In how many ways can the letter of the word PERMUTATIONS can be arranged so that all the vowels come together

## Answer:

Step 1: Count the number of vowels in the word PERMUTATIONS.

Answer: There are 3 vowels in the word PERMUTATIONS (E, U, A).

Step 2: Calculate the number of ways to arrange the vowels (E, U, A).

Answer: The number of ways to arrange the vowels is 3! = 6.

Step 3: Calculate the number of ways to arrange the remaining letters (P, R, M, T, I, N, S).

Answer: The number of ways to arrange the remaining letters is 7! = 5040.

Step 4: Calculate the total number of ways to arrange the letters of the word PERMUTATIONS so that all the vowels come together.

Answer: The total number of ways to arrange the letters of the word PERMUTATIONS so that all the vowels come together is 6 x 5040 = 30240.

## Question:

How many 3-digit numbers can be formed by using the digits 1 to 9 if no digit is repeated?

## Answer:

Answer:

Step 1: The number of possible 3-digit numbers that can be formed using the digits 1 to 9 is equal to the number of permutations of 9 different digits taken 3 at a time.

Step 2: The formula for the number of permutations of n different objects taken r at a time is nPr = n!/(n-r)!

Step 3: Substituting n = 9 and r = 3, we get 9P3 = 9!/(9-3)! = 9!/(6!) = 9x8x7 = 504.

Step 4: Therefore, the number of 3-digit numbers that can be formed using the digits 1 to 9 without repeating any digit is 504.

## Question:

How many words, with or without meaning, can be formed using all the letters of the word EQUATION, using each letter exactly once?

## Answer:

Answer:

- Start by listing all the letters of the word EQUATION: E, Q, U, A, T, I, O, N
- Using each letter exactly once, the maximum number of words that can be formed is 8!
- Therefore, the total number of words, with or without meaning, that can be formed using all the letters of the word EQUATION, using each letter exactly once is 40,320.

## Question:

In how many ways can a letters of the word PERMUTATIONS be arranged if the (i) Words with start with P and end with S (ii) vowels are all together (iii) There are always 4 letter between P and S

## Answer:

(i) The number of ways to arrange the letters of the word PERMUTATIONS where the word starts with P and ends with S is 8!/(2! * 2!) = 20160.

(ii) The number of ways to arrange the letters of the word PERMUTATIONS where the vowels are all together is 8!/(2! * 4!) = 5040.

(iii) The number of ways to arrange the letters of the word PERMUTATIONS where there are always 4 letters between P and S is 8!/(4!) = 20160.