Employees of a certain company are each to receive a unique 7-digit id

One can use slot method to solve this.
7 digit identification code has 7 slots to fill.
_ _ _ _ _ _ _ .
For the first slot there are three possible options. (1,2 and 3 because first digit has to be half of second)
For the second slot there are three possible options. (2,4 and 6 because second digit has to be twice the second)
To fill the first two slots there are only three possible ways. Because if the ,
if the first is 1 the second has to be 2 (only 1 way)
if the first is 2 the second has to be 4 (only 1 way)
if the first is 3 the second has to be 6 (only 1 way) . So total ways to fill first two slots = 3ways
You are now left with 5 digits and 5 slots to fill, which can be filled in 5! ways i.e. 120 ways.

Total no: of valid codes will be therefore : 120 * 3 = 360ways.

Answer : D

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For the first two digits we have 3 variants:
1-2, 2-4, 3-6

For the last 5 digits we have 5 numbers (because we used two numbers) so it will be equal to 5! = 120

3 variants of the first two digits * 120 variants of last 5 digits = 360
Answer is D
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Ans: D

Solution: given the condition first digit can have 3 numbers only 1,2,3 and corresponding second digit will be 2,4,6: now from counting principal.

Every first digit we choose there will be only one possibility for the second place. and first place has 3 possibilities. no repetition is allowed so
: 3 1 5 4 3 2 1
= 360
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Could only be 12XXXXX, 24XXXXX, 36XXXXX. For each of the 3, there are 5! possibilities. Therefore the answer is \(5!*3=5*4*3*2*1*3=20*6*3=120*3=360\) or D

First 2 places can be filled in 3 ways :
1-2, 2-4, 3-6;Last 5 places can be filled in 5!ways = 120
3 ways x 5! ways = 3x120=360
Answer D

MANHATTAN GMAT OFFICIAL SOLUTION:

Valid codes must have a second digit that is exactly twice the first digit. There are three ways to do this with the available digits:

Scenario A: 12XXXXX
Scenario B: 24XXXXX
Scenario C: 36XXXXX

For each of these basic scenarios, there are 5! ways we can shuffle the remaining 5 numbers (represented by X's above).

Thus, the total number of valid codes is 3 × 5! = 3 × 120 = 360.

The correct answer is D.
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Since the second digit is twice the first, the first two digits of a valid code can only be 12, 24, or 36.

For each of these first two digits, the last five digits can be arranged in 5! = 120 ways; thus, there are 3 x 120 = 360 valid codes.

Answer: D
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- we have only 3 ways to choose the first 2 digits (12, 24, 36)
- For the last 5 digits: we have 5! ways to arrange.
=> total ways = 3*5! = 360

Hence the answer is D.

total such pairs first two positions ; 1,2 , 2,4, 3,6
and other 5 positions will be filled in 5! ways ; 120
so 120 * 3 ; 360
IMO D

Possible combinations:-

Case 1. 12xxxxx; 5! ways

Case 2. 24xxxxx; 5! ways


Case 3. 36xxxxx; 5! ways

Total= 3*(5!) is answer.

_ _ _ _ _ _ _
3 6 x 5 x 4 x 3 x 2 x 1
2 4
1 2

The first two digits can be 6 and 3 so the last 5 digits have 5! permutations: 5! = 120
The same applies for the next two rows.

120 + 120 + 120 = 360 because these are all mutually exclusive (i.e. you can't have the first and second digits be 3 and 6 and 2 and 4 at the same time) so you have to add.

Answer is D.
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These types of questions are terribly time consuming.

Employees of a certain company are each to receive a unique 7-digit identification code consisting of the digits 0, 1, 2, 3, 4, 5, and 6 such that no digit is used more than once in any given code. In valid codes, the second digit in the code is exactly twice the first digit. How many valid codes are there?

(A) 42
(B) 120
(C) 210
(D) 360
(E) 840

The only possible numbers for the first digit are 1,2,3. Let's take 1, 2, and 3 each.

1 , 2 (2nd digit), 5 x 4 x 3 x 2 x 1 = 120 possible codes
2, 4 (2nd digit), 5 x 4 x 3 x 2 x 1 = 120 possible codes
3 , 6 (2nd digit), 5 x 4 x 3 x 2 x 1 = 120 possible codes

120 x 3 = 360

D.
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Can someone shed light on how to recognise that these questions will require factorials and slots? That is one issue I'm facing. Is there a marker or anything that can tell me quickly that is the case? Bunuel