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

Question

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!}

Archit3110 · Answer

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

IMO B

aesop · Answer

5!/2! =  5P_3

QuantMadeEasy · Answer

5!/2! = 5P3

B is correct

Luca1111111111111 · Answer

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?

QuantMadeEasy · Answer

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

Luca1111111111111 · Answer

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

pepperoniPizza · Answer

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

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

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