Question

║⊕QUESTION⊕║
MATHS BREAKS MY HEART !!! PLEASE ANSWER THIS QUESTION
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CLASS 11
PERMUTATIONS AND COMBINATIONS
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How many numbers between 3000 and 4000 can be formed from the digits 2, 3, 4, 5, 6 and 7, no digit being repeated in any number?

Answer 1

Answer:

Since you have to form a no. between 3000 and 4000, hence the first digit can only be 3, the second digit can be anything among 2,4,5,6,7 (not 3, because its been used.) So, it can be occupied by any of these 5 numbers.

The third digit can be occupied by any of the remaining 4 numbers.

the fourth digit, by any of the remaining 3. So, by multiplying the no. of ways in which these 4 digits (since there are 4 places in 3000 to 4000) can be occupied, we get, 1 × 5 × 4 × 3 = 60 ways.

Hence, the answer is 60.

I hope you understood :)

Answer 2

Answer:

60

Step-by-step explanation:

Between 3000 and 4000

All the numbers are 4-digited and starts with 3.

So,

Number of ways of filling the thousand place is only one(i.e 3)

Number of filling the hundred place is 5( as 3 is already been used)

Number of filling the tens place is 4

Number of filling the ones place is 3.

So,total number of ways of forming the required number is 1*5*4*3=60..

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