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In how many different ways, the letters of the word ALGEBRA can be arranged in a row if.
(i) The two As are together?
(ii) The two As are not together?
- a)(i) 720, (ii) 1800
- b)(i) 620, (ii) 1600
- c)(i) 780, (ii) 1860
- d)(i) 720, (ii) 1600
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
In how many different ways, the letters of the word ALGEBRA can be ar...
ALGEBRA has seven letters where 2-A, 1-L, 1-G, 1- E, 1-Band 1-R.
(i) Since two A’s are always together, we take both the A's as one letter.
If p is the number of arrangements, then p = 6! = 6 x 5 x 4 x 3 x 2 x 1 = 720.
(ii) Total number of permutations
= 2520
In these permutations, in some permutations, two A's are together while in the rest they are not together Hence, the number of permutations in which two A's are not together in q - p = 2520 - 720 = 1800.
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In how many different ways, the letters of the word ALGEBRA can be ar...
Solution:
(i) The two As are together:
To solve this, we can treat the two As as a single entity. So, we have 6 letters (A, L, G, E, B, R) along with the combined entity 'AA'.
The number of arrangements of these 7 entities is 7!, but since the two As are identical, we need to divide by 2! to account for the repetition. Therefore, the number of arrangements when the two As are together is:
7!/2! = 7 * 6! / 2 * 1 = 7 * 720 = 5040
(ii) The two As are not together:
To solve this, we can consider the total number of arrangements of the word ALGEBRA and subtract the number of arrangements where the two As are together.
The total number of arrangements of the 7 letters in the word ALGEBRA is 7!.
From part (i), we know that the number of arrangements where the two As are together is 5040.
Therefore, the number of arrangements where the two As are not together is:
7! - 5040 = 5040 - 5040 = 0
So, there are no arrangements of the word ALGEBRA where the two As are not together.
Therefore, the correct answer is option 'A' (i) 720, (ii) 1800.
(i) The two As are together:
To solve this, we can treat the two As as a single entity. So, we have 6 letters (A, L, G, E, B, R) along with the combined entity 'AA'.
The number of arrangements of these 7 entities is 7!, but since the two As are identical, we need to divide by 2! to account for the repetition. Therefore, the number of arrangements when the two As are together is:
7!/2! = 7 * 6! / 2 * 1 = 7 * 720 = 5040
(ii) The two As are not together:
To solve this, we can consider the total number of arrangements of the word ALGEBRA and subtract the number of arrangements where the two As are together.
The total number of arrangements of the 7 letters in the word ALGEBRA is 7!.
From part (i), we know that the number of arrangements where the two As are together is 5040.
Therefore, the number of arrangements where the two As are not together is:
7! - 5040 = 5040 - 5040 = 0
So, there are no arrangements of the word ALGEBRA where the two As are not together.
Therefore, the correct answer is option 'A' (i) 720, (ii) 1800.
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In how many different ways, the letters of the word ALGEBRA can be arranged in a row if.(i) The two As are together?(ii) The two As are not together?a)(i) 720, (ii) 1800b)(i) 620, (ii) 1600c)(i) 780, (ii) 1860d)(i) 720, (ii) 1600Correct answer is option 'A'. Can you explain this answer?
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