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:

1. Start by listing all the letters of the word EQUATION: E, Q, U, A, T, I, O, N
2. Using each letter exactly once, the maximum number of words that can be formed is 8!
3. 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.