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Math Expert V
Joined: 02 Sep 2009
Posts: 57155
How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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6 00:00

Difficulty:   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 number 11235 only once?

(A) $$5!$$

(B) $$5P_3$$

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

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

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

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

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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.
Intern  B
Joined: 20 Jun 2013
Posts: 15
Location: India
WE: Consulting (Health Care)
Re: How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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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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GPA: 3.97
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Re: How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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5!/2! = 5P3

B is correct
Manager  B
Joined: 20 Apr 2019
Posts: 81
Re: How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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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?
Manager  S
Joined: 28 Feb 2014
Posts: 137
Location: India
Concentration: General Management, International Business
GPA: 3.97
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Re: How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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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?

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
Manager  B
Joined: 20 Apr 2019
Posts: 81
Re: How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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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
Intern  B
Joined: 03 Jun 2019
Posts: 4
How many unequal 5-digit numbers can be formed using each digit of the  [#permalink]

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1
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. How many unequal 5-digit numbers can be formed using each digit of the   [#permalink] 06 Aug 2019, 03:21
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# How many unequal 5-digit numbers can be formed using each digit of the

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