SATHEE: Permutations And Combinations Question 310

Permutations And Combinations Question 310

Question: How many numbers lying between 999 and 10000 can be formed with the help of the digit 0,2,3,6,7,8 when the digits are not to be repeated [AMU 2005]

Options:

A) 100

B) 200

C) 300

D) 400

Show Answer

Answer:

Correct Answer: C

Solution:

The numbers between 999 and 10000 are of four digit numbers. The four digit numbers formed by digits 0, 2, 3, 6, 7, 8 are $ ^{6}P_4=360 $ . But here those numbers are also involved which begin from 0. So we take those numbers as three digit numbers. Taking initial digit 0, the number of ways to fill remaining 3 places from five digits 2, 3, 6, 7, 8 are $ ^{5}P_3=60 $ So the required numbers = 360-60 = 300.

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Permutations And Combinations Question 310

Question: How many numbers lying between 999 and 10000 can be formed with the help of the digit 0,2,3,6,7,8 when the digits are not to be repeated [AMU 2005]

Options:

Answer:

Solution:

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