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? Answer: (A) Explanation: The word “APPLE” has 5 letters in which “P” comes twice. Quiz of this Question
(A) 1024, 60
(B) 60, 1024
(C) 1024, 1024
(D) 240, 1024
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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? Answer: (A) Explanation: We have to pick 3 men from 6 available men and 3 ladies from 7 available ladies. Quiz of this Question
(A) 700
(B) 720
(C) 120
(D) 500
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? 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). Hence, the number of ways = 20!* 21P18 Quiz of this Question
(A) 20!* 20C18
(B) 20!* 20P18
(C) 20!* 21C18
(D) 20!* 21P18