SATHEE: Permutations And Combinations Question 283

Permutations And Combinations Question 283

Question: How many numbers, lying between 99 and 1000 be made from the digits 2, 3, 7, 0, 8, 6 when the digits occur only once in each number [MP PET 1984]

Options:

A) 100

B) 90

C) 120

D) 80

Show Answer

Answer:

Correct Answer: A

Solution:

Required number of ways = $ ^{6}P_3{{-}^{5}}P_2=120-20=100 $ . {Since between 99 and 1000, the numbers are of 3 digits, but those numbers should be omit in which ?0? comes at hundred place}.

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Permutations And Combinations Question 283

Question: How many numbers, lying between 99 and 1000 be made from the digits 2, 3, 7, 0, 8, 6 when the digits occur only once in each number [MP PET 1984]

Options:

Answer:

Solution:

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