How many 5-digit numbers can be formed from the digits 1, 2, 3, 4, 5, 6, 7, 8, 9?

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Math Expert

Joined: 02 Sep 2009

Posts: 87762

How many 5-digit numbers can be formed from the digits 2, 3, 5, 6, 8, [#permalink]

  01 Mar 2021, 04:46

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Question Stats:

87% (00:42) correct 13% (00:39) wrong based on 45 sessions

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How many 5-digit numbers can be formed from the digits 2, 3, 5, 6, 8, 9 if no digit can be used more than once in a number ?

A. 120
B. 240
C. 360
D. 720
E. 1080

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Affiliations: CrackVerbal

Posts: 4988

Location: India

Re: How many 5-digit numbers can be formed from the digits 2, 3, 5, 6, 8, [#permalink]

  01 Mar 2021, 05:13

Basically the question is asking how many 5 digit numbers can be formed without repetition.

If we do by the box method, then the 1st place can be filled in 6 ways, second in 5, 3rd in 4 4th in 3 and the last in 2 ways.

Total number of ways = 6 * 5 * 4 * 3 * 2 = 720

In terms of permutations, we need to arrange 5 digit out of 6 = 6P5 = 6! / (6 - 5)! = 720 ways

In terms of combinations, first choose 5 numbers out of 6 in 6C5 ways and then arrange these 5 digits in 5! ways.

Total arrangements = 6C5 * 5! = 6 * 5! = 720

Option D

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How many 5-digit numbers can be formed from the digits 2, 3, 5, 6, 8, [#permalink]

  01 Mar 2021, 06:28

Bunuel wrote:

How many 5-digit numbers can be formed from the digits 2, 3, 5, 6, 8, 9 if no digit can be used more than once in a number ?

A. 120
B. 240
C. 360
D. 720
E. 1080

We have 5 empty places like this

- - - - -

We have 6 digits to use and the digit can NOT be repeated

Method-1

So we can fill the first place in 6 ways

6 - - - -

So we can fill the Second place in 5 ways (can not use the used digit at first place)

6 * 5 - - -

So we can fill the Third place in 4 ways (can not use the used digits at previous places)

6 * 5 * 4 - -

So we can fill the Forth place in 4 ways (can not use the used digits at previous places)

6 * 5 * 4 * 3 -

So we can fill the Fifth place in 4 ways (can not use the used digits at previous places)

6 * 5 * 4 * 3 * 2

So total numbers =

720

ANswer: Option D

Method-2

Step-1: Select 5 out of 6 digits that need to be used to make a five digit number in

\(^6C_5\)

ways
Step-2: Arrange those selected digits at 5 places in

5!

ways

Total Such numbers =

\(^6C_5*5! = 720\)

ANswer: Option D

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Re: How many 5-digit numbers can be formed from the digits 2, 3, 5, 6, 8, [#permalink]

  01 Mar 2021, 08:02

Total numbers: 6 [2, 3, 5, 6, 8, 9]

Condition: Repetition not allowed

Five-digit number: _, _, _, _, _

For first place from left, we have the choice of 6 numbers
For second place from left, we have the choice of 5 numbers
For third place from left, we have the choice of 4 numbers
For fourth place from left, we have the choice of 3 numbers
For fifth place from left, we have the choice of 2 numbers

=> 6 * 5 * 4 * 3 * 2 = 720

Answer D
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Re: How many 5-digit numbers can be formed from the digits 2, 3, 5, 6, 8, [#permalink]

01 Mar 2021, 08:02

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