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How many unequal 5-digit numbers can be formed using each digit of the

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How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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New post 26 Jul 2019, 00:32
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  45% (medium)

Question Stats:

54% (01:13) correct 46% (01:17) wrong based on 69 sessions

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How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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New post Updated on: 27 Jul 2019, 06:39
Bunuel wrote:
How many unequal 5-digit numbers can be formed using each digit of the number 11235 only once?

(A) \(5!\)

(B) \(5P_3\)

(C) \(\frac{5C_5}{2!}\)

(D) \(\frac{5P_5}{2!*3!}\)

(E) \(\frac{5C_5}{2!*3!}\)


total ways to arrange 11235 ; 5!/2! =60 i.e \(5P_3\)

IMO B
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Originally posted by Archit3110 on 26 Jul 2019, 07:23.
Last edited by Archit3110 on 27 Jul 2019, 06:39, edited 1 time in total.
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Re: How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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New post 26 Jul 2019, 10:35
Archit3110 wrote:
Bunuel wrote:
How many unequal 5-digit numbers can be formed using each digit of the number 11235 only once?

(A) \(5!\)

(B) \(5P_3\)

(C) \(\frac{5C_5}{2!}\)

(D) \(\frac{5P_5}{2!*3!}\)

(E) \(\frac{5C_5}{2!*3!}\)


total ways to arrange 11235 ; 5!/2! or say \(\frac{5C_5}{2!}\)

IMO C

5!/2! = \(5P_3\)
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Re: How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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New post 26 Jul 2019, 12:02
5!/2! = 5P3

B is correct
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Re: How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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New post 28 Jul 2019, 08:11
Bunuel wrote:
How many unequal 5-digit numbers can be formed using each digit of the number 11235 only once?

(A) \(5!\)

(B) \(5P_3\)

(C) \(\frac{5C_5}{2!}\)

(D) \(\frac{5P_5}{2!*3!}\)

(E) \(\frac{5C_5}{2!*3!}\)

Can someone please explain:
I thought it is 4*3*2*1*4 = 96 because it is asked for an unequal 5-digit number meaning that the 2 cannot be the last digit?
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Re: How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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New post 28 Jul 2019, 09:43
Luca1111111111111 wrote:
Bunuel wrote:
How many unequal 5-digit numbers can be formed using each digit of the number 11235 only once?

(A) \(5!\)

(B) \(5P_3\)

(C) \(\frac{5C_5}{2!}\)

(D) \(\frac{5P_5}{2!*3!}\)

(E) \(\frac{5C_5}{2!*3!}\)

Can someone please explain:
I thought it is 4*3*2*1*4 = 96 because it is asked for an unequal 5-digit number meaning that the 2 cannot be the last digit?


5 digits can be arranged in 5! ways, but 2 digits are same so divide by 2! (as 2 digits can be rearranged within themselves)
5!/2! = 5P3
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Re: How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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New post 01 Aug 2019, 05:06
kapil1 wrote:
Luca1111111111111 wrote:
Bunuel wrote:
How many unequal 5-digit numbers can be formed using each digit of the number 11235 only once?

(A) \(5!\)

(B) \(5P_3\)

(C) \(\frac{5C_5}{2!}\)

(D) \(\frac{5P_5}{2!*3!}\)

(E) \(\frac{5C_5}{2!*3!}\)

Can someone please explain:
I thought it is 4*3*2*1*4 = 96 because it is asked for an unequal 5-digit number meaning that the 2 cannot be the last digit?


5 digits can be arranged in 5! ways, but 2 digits are same so divide by 2! (as 2 digits can be rearranged within themselves)
5!/2! = 5P3

Sorry but this does not answer my question. I would like to know if we do not have to specifically change something in the calculation because we are looking for an unequal 5-digit number, meaning that we need to exclude the 2 as a possible last digit.
If we would change the question to ask for a 5-digit number with given values of 1,1,2,3,5 then I would assume that it is also 5P3 which cannot be as one time we look for an unequal number and another time for a number without restrictions
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How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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New post 06 Aug 2019, 03:21
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Luca1111111111111 wrote:
Bunuel wrote:
How many unequal 5-digit numbers can be formed using each digit of the number 11235 only once?

(A) \(5!\)

(B) \(5P_3\)

(C) \(\frac{5C_5}{2!}\)

(D) \(\frac{5P_5}{2!*3!}\)

(E) \(\frac{5C_5}{2!*3!}\)

Can someone please explain:
I thought it is 4*3*2*1*4 = 96 because it is asked for an unequal 5-digit number meaning that the 2 cannot be the last digit?


I think you're confusing unequal with uneven? (as in odd and even). The question ask for unequal (ie different), so the 2 can be the last digit.
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How many unequal 5-digit numbers can be formed using each digit of the   [#permalink] 06 Aug 2019, 03:21
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