How many words can be made from the word Apple using all the alphabets without repetition respectively?

I think it should be 5^5 as AAAAA and PPPPP...LLLLL EEEEE is also a case. There is no mention of with or without repetition nor words with meaning.

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How many words can be made from the word “APPLE” using all the alphabets with repetition and without repetition respectively?
(A) 1024, 60
(B) 60, 1024
(C) 1024, 1024
(D) 240, 1024

Answer: (A)

Explanation: The word “APPLE” has 5 letters in which “P” comes twice.

If repetition is allowed, the number of words we can form = 4*4*4*4*4 = 1024.(This is because, when repetition is allowed, we can put any of the four unique alphabets at each of the five positions.)If repetition is not allowed, the number of words we can form = 5!/2! = 60. (This is because “P” comes twice.)

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How many ways a 6 member team can be formed having 3 men and 3 ladies from a group of 6 men and 7 ladies?
(A) 700
(B) 720
(C) 120
(D) 500

Answer: (A)

Explanation: We have to pick 3 men from 6 available men and 3 ladies from 7 available ladies.

Required number of ways = 6C3 * 7C3 = 20 * 35 = 700.

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In how many ways can an interview panel of 3 members be formed from 3 engineers, 2 psychologists and 3 managers if at least 1 engineer must be included?
(A) 30
(B) 15
(C) 46
(D) 45

Answer: (C)

Explanation: The interview panel of 3 members can be formed in 3 ways by selecting 1 engineer and 2 other professionals, 2 engineers and 1 other professionals and all 3 engineers.

1 engineer out of 3 engineers and 2 other professionals out of 5 professionals can be selected as
= 3C1 * 5C2 = 3 * 10 = 30 ways.
2 engineers out of 3 engineers and 1 other professional out of 5 professionals can be selected as
= 3C2 * 5C1 = 3 * 5 = 15 ways.
3 engineers out of 3 engineers and 0 other professional out of 5 professionals can be selected as
= 3C3 * 5C0 = 1 way.

Hence, total number of ways = 30 + 15 + 1 = 46 ways.

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How many 4-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6 and 7 which are divisible by 5 when none of the digits are repeated?
(A) 120
(B) 35
(C) 24
(D) 720

Answer: (A)

Explanation: A number is divisible by 5 if and only if its last digit is either 5 or 0. But, 0 is not available here. So, we have to fix 5 as a last digit of 4-digit number and fill 3 places with remaining 6 digits.

Number of ways to choose 3 digits = 6C3 = 20.Number of ways to arrange the chosen digits = 3!

Hence, total number of required ways = 6C3 * 3! = 6P3 = 120.

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In how many ways can 20 boys and 18 girls make a queue such that no two girls are together?
(A) 20!* 20C18
(B) 20!* 20P18
(C) 20!* 21C18
(D) 20!* 21P18

Answer: (D)

Explanation: The boys will be arranged in 20! ways. Now, there are a total of 21 possible places available between boys such that no 2 girls can be placed together (alternate sequence of boys and girls, starting and ending positions for girls).

Therefore, the 18 girls can stand at these 21 places only.

Hence, the number of ways = 20!* 21P18

Option (D) is correct.

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