failatmath
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 ?
A. 120
B. 180
C. 288
D. 360
E. 720
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 into cases:
If we form a number without 2 and without 5, then we must use the numbers 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
(C) is our answer.