Find the total number of different permutations of all the letters of the word:
a. ADA
b. SAYA
c. NANA
d. AGAMA
a. ADA
The word "ADA" has 3 distinct letters (A, D, and A). To find the total number of different permutations, we can use the formula for permutations of n objects with repetition, which is n!.
In this case, n = 3. So, the total number of different permutations of the letters in "ADA" is 3! = 3 × 2 × 1 = 6.
Therefore, there are 6 different permutations of the letters in "ADA".
b. SAYA
The word "SAYA" has 4 distinct letters (S, A, Y, and A). Again, using the formula for permutations of n objects with repetition, the total number of different permutations is 4!.
In this case, n = 4. So, the total number of different permutations of the letters in "SAYA" is 4! = 4 × 3 × 2 × 1 = 24.
Therefore, there are 24 different permutations of the letters in "SAYA".
c. NANA
The word "NANA" has 4 letters, but only 2 distinct letters (N and A). To find the total number of different permutations, we once again use the formula for permutations of n objects with repetition, but this time we divide by the factorial of the number of times each repeated letter appears.
In this case, n = 4 and there are two A's. So, the total number of different permutations of the letters in "NANA" is 4! / (2!) = (4 × 3 × 2 × 1) / (2 × 1) = 12.
Therefore, there are 12 different permutations of the letters in "NANA".
d. AGAMA
The word "AGAMA" has 5 letters, but only 3 distinct letters (A, G, and M). Similar to the previous cases, we use the formula for permutations of n objects with repetition, but this time we also divide by the factorials of the number of times each repeated letter appears.
In this case, n = 5, there are two A's, and one each of G and M. So, the total number of different permutations of the letters in "AGAMA" is 5! / (2! × 1! × 1!) = (5 × 4 × 3 × 2 × 1) / (2 × 1 × 1 × 1) = 60.
Therefore, there are 60 different permutations of the letters in "AGAMA".
To find the total number of different permutations of all the letters of a word, we need to calculate the factorial of the number of letters in the word.
a. ADA:
The word ADA has 3 letters. To find the total number of different permutations, we calculate 3 factorial (3!).
3! = 3 x 2 x 1 = 6
So, there are 6 different permutations of the letters in the word ADA.
b. SAYA:
The word SAYA has 4 letters. To find the total number of different permutations, we calculate 4 factorial (4!).
4! = 4 x 3 x 2 x 1 = 24
So, there are 24 different permutations of the letters in the word SAYA.
c. NANA:
The word NANA has 4 letters. To find the total number of different permutations, we calculate 4 factorial (4!).
4! = 4 x 3 x 2 x 1 = 24
So, there are 24 different permutations of the letters in the word NANA.
d. AGAMA:
The word AGAMA has 5 letters. To find the total number of different permutations, we calculate 5 factorial (5!).
5! = 5 x 4 x 3 x 2 x 1 = 120
So, there are 120 different permutations of the letters in the word AGAMA.