How many 3-digit numbers can be formed using the digits 1 2 3 4 5 6 7 8 9 with repetition

Solution : (i) When repetition of digits is allowed:
No. of ways of choosing firsy digits = 5
No. of ways of choosing second digit = 5
No. of ways of choosing third digit = 5
Therefore, total possible numbers `= 5 xx 5 xx 5 = 125`
(ii) When repetition of digits is not allowed:
No. of ways of choosing first digit = 5
No. of ways of choosing second digit = 4
No. of ways of choosing thrid digit = 3
Total possible numbers `= 5 xx 4 xx 3 = 60`.

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Solution

Given 3 digit numbers that can be formed using the digits from 1 to 9.Required number of 3-digit numbers= arranging 3 digits with the total number of 9 digits.= 9P3=(9×8×7)=504.Hence, the required number of numbers =504.

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How many possible 3 digit numbers can be made using the digits 604?

There are 4 possible numbers if the digits are not repeated; 18 if they are. Those are 3-digit numbers, assuming that zero would not be a leading digit. If zero is allowed for a leading digit, then you can have 6 for the non repeated, and 27 if repetition.

Thea E.

asked • 01/24/16

how many three digit numbers can be made with the digits 1,2,3,4,5,6 and 7 if a) repetitions are allowed b) no digits is to be repeated in a number?

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1 Expert Answer

How many 3-digit numbers can be formed using the digits 1 2 3 4 5 6 7 8 9 with repetition

Hi Thea

With 7 numbers, we can create three digit numbers like this.

a)  With repetitions would suggest with replacement.

We'd have 7 options for the first digit, 7 options for the 2nd digit and 7 choices for the 3rd digit

73 = 343 combinations

b) Without repeated numbers, would suggest without replacement

We'd have 7 options for the first digit, 6 options for the 2nd digit and 5 choices for the 3rd digit

7•6•5 = 210 combinations

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How many 3

(i) repetition of the digits is allowed? Solution: Answer: 125.

How many 3

Note: The number of 3-digit numbers formed using the digits 1, 2, 3, 4 and 5 when repetition is not allowed is equivalent to the number of 3 Letter permutations of 5 distinct letter = 5P3=5! (5−3)! =5! 2!

How many three digit numbers can be formed using the digits 1 2 3 4 5 if a repetition of the digits is allowed 2 B repetition of the digits is not allowed?

so 60(ans.)

How many 3

∴ Required number of numbers = (1 x 5 x 4) = 20.