A five-digit code for certain locks uses the digits 0,1,2,3,4,5,6,7,8,

Question

Permutation and Combination - Practice Question #3

3-  A five-digit code for certain locks uses the digits 0,1,2,3,4,5,6,7,8,9 according to the following pattern. The first digit must be 2 or 4, 3rd digit must be a multiple of 3 and no two consecutive digits should be same. How many different codes are possible.

Options

a)  2,016
 b)  3,888
 c)  4,608
 d)  5,184
 e)  6,236

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To solve Question 2:    Question 2

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EgmatQuantExpert · Answer

Solution

Given:

• A five-digit code uses the digits 0,1,2,3,4,5,6,7,8,9.
• The first digit must be 2 or 4
• 3rd digit must be a multiple of 3
• No two consecutive digits should be same

To find:  
 • The number of different codes possible.

Approach and Working:

Let the 5-digit number is abcde.

• Ways to fill ‘a’ = 2(2 or 4)

• Ways to fill ‘c’ = 4 (0,3,6,9)

• Hence, we filled two different digits in ‘a’ and ‘c’. 
 o So, out of 10 digits, b cannot take 2 digits. 
o Thus, ‘b’ can be filled in 8 ways.

• Now, ‘d’ can be filled in 9 ways as similar digit cannot be in ‘c’ and ‘d’.

• Similarly, ‘e’ can be filled in 9 ways.

Thus, total ways to fill= 2*8*4*9*9= 5184

Hence, the correct answer is option d.

Answer: d

Hero8888 · Answer

A five-digit code for certain locks uses the digits 0,1,2,3,4,5,6,7,8,9 according to the following pattern. The first digit must be 2 or 4, 3rd digit must be a multiple of 3 and no two consecutive digits should be same. How many different codes are possible.

1st cell - 2 options (2,4) - the first digit must be 2 or 4
3rd cell - 4 options (0,3,6,9) - multiples of 3
2nd cell - 8 options - can not be surrounded by the same digits that are located in cells 1 and 3, so we have to pick from 10-2=8
4th cell - 9 options
5th cell - 9 options

2*8*4*9*9 = 5,184

Answer D

thefibonacci · Answer

code is of the form abcde
let's fill the constraints first:
a can be filled in 2 ways (2 or 4) 
c can be filled in 4 ways (0,3,6, or 9)
b can be filled in 8 ways (can't be same as a or c)
d can be filled in 9 ways (can't be same as c)
e can be filled in 9 ways (can't be same as d)
total = 2*4*8*9*9 = 5184.

raks85 · Answer

IMO it should be 2*8*3*9*9= 3888 (B)
I do not understand how 0 can be considered a multiple of 3. In that way 0 is multiple of each and every number.

EgmatQuantExpert · Answer

Hey  raks85 ,

Yes, 0 can be written as a multiple of any integer.
Regards,
Ashutosh
e-GMAT

EgmatQuantExpert · Answer

Hello everyone,

Thanks for participation.

We have posted the official answer to the question.

Regards,
Ashutosh
e-GMAT

PearlRay · Answer

egmat

I have a doubt -in digit abcde d can also take the value of e right? d is sandwiched between c and e ( as b is sandwiched between a and c) . de is also a consecutive pair- so shouldn't the no of ways to fill d be 8? and that of e be 9

Kem12 · Answer

I'm also not clear as why d can be 9 ways since it cannot be the same as c or e and there are no restrictions on d. Are c-d and d-e not Consecutive pairs as well? Bunuel, please help me understand my mistake. Thanks.

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Posted from my mobile device

oceantux · Answer

I had the exact same doubt. One possible explanation is :

Going as per solution, If the code is abcde, then,

i) First we choose, "a" - this has 2 ways of being chosen (2,4).
ii) then we choose "c" - this has 4 ways of being chosen (yes, 0 is a multiple- 0,3,6,9).
iii) next, we lock on "b" - this has 8 ways of being chosen (SINCE "b" cannot be same as "a" or "c"); 
moving forward, 
iv) to choose "d" - we have 9 possible ways here because "d" cannot be same as "c", AND HERE we have TO realize that "e" has not yet been chosen...
hence, now that "d" is locked, 
v) we move forward to choose "e" - 9 ways to choose "e" because "e" cannot be same as "d".

with this condition, both "d" and "e" will be unique.

=> Final answer is 2x8x4x9x9 = 5184

I hope my reasoning is clear and unambiguous.

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A five-digit code for certain locks uses the digits 0,1,2,3,4,5,6,7,8,

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