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# How many five-digit numbers can be formed using the digits 5

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Joined: 07 Jun 2014
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How many five-digit numbers can be formed using the digits 5 [#permalink]  30 Jul 2018, 10:46
Expert's post
00:00

Question Stats:

71% (00:34) correct 28% (01:31) wrong based on 14 sessions
How many five-digit numbers can be formed using the digits 5, 6, 7, 8, 9, 0 if no digits can be repeated?

(A) 64
(B) 120
(C) 240
(D) 600
(E) 720
[Reveal] Spoiler: OA

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Sandy
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Re: How many five-digit numbers can be formed using the digits 5 [#permalink]  05 Aug 2018, 21:26
sandy wrote:
How many five-digit numbers can be formed using the digits 5, 6, 7, 8, 9, 0 if no digits can be repeated?

(A) 64
(B) 120
(C) 240
(D) 600
(E) 720

GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4856
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 102

Kudos [?]: 1731 [0], given: 397

Re: How many five-digit numbers can be formed using the digits 5 [#permalink]  06 Aug 2018, 06:26
Expert's post
We have 5 digit number represented by 5 blank spaces _ _ _ _ _ . We have 6 numbers to fill these blank spaces.

The first blank can be any digit other than 0. So 5 ways to fill this blank.

The second blank can be filled again in 5 ways i.e. all the digits except for the digit used in first blank.

Third blank is 4 ways

Fourth blank is 3 ways

And Fifth blank is 2 ways.

So the total number of combinations are $$5 \times 5 \times 4 \times 3 \times 2=600$$
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GRE Prep Club Legend
Joined: 07 Jun 2014
Posts: 4856
GRE 1: Q167 V156
WE: Business Development (Energy and Utilities)
Followers: 102

Kudos [?]: 1731 [0], given: 397

Re: How many five-digit numbers can be formed using the digits 5 [#permalink]  07 Aug 2018, 06:13
Expert's post
Explanation

This problem relies on the fundamental counting principle, which says that the total number of ways for something to happen is the product of the number of options for each individual choice.

The problem asks how many five-digit numbers can be created from the digits 5, 6, 7, 8, 9, and 0. For the first digit, there are only five options (5, 6, 7, 8, and 9) because a five-digit number must start with a non-zero integer.

For the second digit, there are 5 choices again, because now zero can be used but one of the other numbers has already been used, and numbers cannot be repeated. For the third
number, there are 4 choices, for the fourth number there are 3 choices, and for the fifth number there are 2 choices. Thus, the total number of choices is (5)(5)(4)(3)(2) = 600.

Alternatively, use the same logic and realize there are 5 choices for the first digit. (Separate out the first step because you have to remove the zero from consideration.) The remaining five digits all have an equal chance of being chosen, so choose four out of the remaining five digits to complete the number. The number of ways in which this second step can be accomplished is $$\frac{5!}{1!}= (5)(4)(3)(2)$$.

Thus, the total number of choices is again equal to (5)(5)(4)(3)(2) = 600.
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Re: How many five-digit numbers can be formed using the digits 5 [#permalink]  02 Jan 2019, 22:38
Expert's post
Lets make this easier.
we know that the total number of combinations is 6!
but we are limited by not being able to put a 0 as the beginning digit.

we know then that we have 6! - (all combinations with 0 as the beginning digit)

to find out all the combinations where the number begins with 0, is just 5!.

just subtract 6! - 5! and you have 720-120=600.. D is the answer.

in case you get confused how to calculuate all combinations with 0 as the beginning digit, keep reading...

Just think

line up the possible number combinations

we know that zero is the beginning digit, so there is only one combination for that

1

but we have 5 number choices remaining for the next digit
1*5
we have 4 number choices remaining for the next digit.
1*5*4

and so on and so on until we have:
1*5*4*3*2*1=120

hope that helped!
Re: How many five-digit numbers can be formed using the digits 5   [#permalink] 02 Jan 2019, 22:38
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