How Many 4 Digit Numbers can be Formed Using the Digits 1, 2, 5, 6, 8 and 9 Without Repetition, So that 2 and 5 Never Come Together ? GMAT Problem Solving

Question: How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8 and 9 without repetition, so that 2 and 5 never come together ?

1. 120
2. 180
3. 288
4. 360
5. 720

Correct Answer: C

Solution and Explanation:
Approach Solution 1:
Problems that ask us to arrange objects under specific constraints can often be split into smaller, simpler problems.

We'll look for such a division, a Logical approach.

We'll split it into cases:

If we form a number without 2 and without 5, then we must use the number 1,6,8,9.

This gives 4! = 24 options.

If we pick only one of 2,5 and another 3 of the remaining numbers this gives 2C1 * 4C3 = 8 options.

This gives 8 * 4! = 192 arrangements (all of which are good as 2,5 are not together)

If we pick 2, 5, and another 2 of the remaining numbers this gives 2C2 * 4C2 = 6 options.

How many 'bad' options, that is options such that 2 and 5 are together, are there?

3! * 2: 2-5-x-y, 2-5-y-x, x-2-5-y, y-2-5-x, x-y-2-5, y-x-2-5 and all of these with the numbers 2,5 reversed.

This is essentially arranging 3 objects: x,y, and a '2-5' clump.

So - we have (4! - 3!*2) = 12 arrangements per option times 6 options giving 72 arrangements.

24 + 192 + 72 = 288

Approach Solution 2:

Total numbers = 6 . (1,2,5,6,8,9)

Total ways = 6*5*4*3 = 360 . (any 6 numbers can take the first slot, any 5 in the second slot ...)

Restriction - 2 and 5 can't be placed right next to each other.

X= 2
Y= 5

First scenario: XY43 = 1*1*4*3 = 12 . Multiply it by 2 since X and Y can switch positions too. = 24

Second scenario: 4XY3 = 4*1*1*3 = 12 Multiply it by 2 since X and Y can switch positions too. = 24

Third scenario: 43XY = 4*3*1*1 = 12 . Multiply it by 2 since X and Y can switch positions too. = 24

Total - unwanted = 360 -24 -24 - 24 = 288

Approach Solution 3:
There are 6P4(6!/(6-4)!= 6*5*4*3*2/2= 360 permutations of 4 digit numbers from the given 6 numbers.

Permutation having adjacent 2 and 5 are of the form:
25ab, where a is 1 of 4, b 1 of 3= 4*3= 12 permutations
52ab, where a is 1 of 4, b 1 of 3= 4*3= 12 permutations
a25b, where a is 1 of 4, b 1 of 3= 4*3= 12 permutations
a52b, where a is 1 of 4, b 1 of 3= 4*3= 12 permutations
ab25, where a is 1 of 4, b 1 of 3= 4*3= 12 permutations
ab52, where a is 1 of 4, b 1 of 3= 4*3= 12 permutations

A total of 6*12= 72 permutations are not allowed.

Therefore, there are 360-72= 288 permutationsof 4 digit numbers from the given set window having 2 and 5 adjacent in them. 

“How many 4 digit numbers can be formed using the digits 1, 2, 5, 6, 8 and 9 without repetition, so that 2 and 5 never come together ?”- is a topic of the GMAT Quantitative reasoning section of GMAT. To solve GMAT Problem Solving questions a student must have knowledge about a good amount of qualitative skills. GMAT Quant practice papers improve the mathematical knowledge of the candidates as it represents multiple sorts of quantitative problems.

 

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