If each digit in the set A = {1, 2, 3, 4, 5} is used exactly once, in

Question

If each digit in the set A = {1, 2, 3, 4, 5} is used exactly once, in how many ways can the 5 digits be arranged?

A. 6
B. 24
C. 72
D. 96
E. 120

However, I came up with 112. Here is how I solved it. Total arrangements can be 5x4x3x2x1 = 120

2& 5 will be adjacent with only these arrangements:
2,5,1,3,4
1,2,5,3,4
1,3,2,5,4
1,3,4,2,5
5,2,1,3,4
1,5,2,3,4
1,3,5,2,4
1,3,4,5,2

120 - 8 = 112.

Where did I go wrong?

gurpreetsingh · Answer

The answer should be = total arrangements - arrangements when 2 and 5 are together.

= 5! - 4! * 2! = 120 -48 = 72

You do not have to count

take 1,3,4,(2,5) here there are 4 entities whose number of arrangements are 4! and 2! is the arrangement between 2 and 5

ENGRTOMBA2018 · Answer

Use the slot method for allthe possible arrangments: _ _ _ _ _ : we have 5 options for the 1st slot, 4 for the 2nd, 3 for 3rd , 2 for 4th and 1 for the 5th, giving the total number of arrangements = 5*4*3*2*1 = 120, E is the correct answer.

GMATinsight · Answer

Total ways of arrangement can be explained by the following steps

- - - - -
5 blank spaces to be filled by each of 5 digits

First Place can be filled in 5 ways (any digit out of available 5 digits)
Second Place can be filled in 4 ways (any digit out of Remaining available 4 digits because one digit has been used already)
Third Place can be filled in 3 ways (any digit out of Remaining available 3 digits because Two digits have been used already)
Forth Place can be filled in 2 ways (any digit out of Remaining available 2 digits because Three digits have been used already)
Fifth Place can be filled in 1 ways (any digit out of Remaining available 1 digits because Four digits have been used already)

Total Arrangements = 5 x 4 x 3 x 2 x 1 = 5! = 120

Answer: Option E

TimeTraveller · Answer

A = {1,2,3,4,5}

Let X1-X2-X3-X4-X5 be the slot to be filled.

X1 slot can be filled by any of the 5 elements of A in 5 ways.
X2 slot can be filled by any of the 4 elements of A in 4 ways.
X3 slot can be filled by any of the 3 elements of A in 3 ways.
X4 slot can be filled by any of the 2 elements of A in 2 ways.
X5 slot can be filled by any of the 5 elements of A in 1 way.

Therefore, 5 digits can be arranged in 5*4*3*2*1 = 120 ways. Ans (E).

reto · Answer

This is combinations without repetition, single source and order does matter. Btw. most questions in GMAT however involve combinations where order doesn't matter.

Here it's simply 5! = 120

You should know the value of 5! by heart.

MBAHopeful99 · Answer

If a number is only used once, the total number of digits available is reduced by one each time.

5*4*3*2*1 = 5! = 120

Bunuel · Answer

800score Official Solution:

There are 5! = 120 possible arrangements of the five digits with no restrictions. Every time an item is selected, it is reduced from the pool. (5)(4)(3)(2)(1) = 120.

sony1000 · Answer

In how many ways can the 5 digits be arranged if then numbers in the set A could be used more than once ?

davedekoos · Answer

In that case, then you could take the same approach, but you would not reduce the available numbers for each "slot" after each step.

5 choices for the first slot
5 choices for the second slot, 
5 choices for the third slot, 
5 choices for the fourth slot,
and 5 choices for the fifth slot

5*5*5*5*5 = 5^5 = 3125

bumpbot · Answer

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