Question

Five numbers A, B, C, D and E are to be arranged in an array in such a manner that they have a common prime factor between two consecutive numbers. These integers are such that:

A has a prime factor P.
B has two prime factors Q and R.
C has two prime factors Q and S.
D has two prime factors P and S.
E has two prime factors P and R.

If B must be arranged at one end in the array, in how many ways can the other four numbers be arranged?

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

B can only sit next to C or E; branch on each choice and follow the forced chain of shared primes to the end, then add up the valid completions.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
B is now pinned down at one end of the row of five numbers. We must count how many different orders the remaining four numbers, A, C, D and E, can take up in the other four spots, while keeping every pair of side-by-side numbers sharing a common prime factor.

Step 2: Key Formula or Approach:
Since B sits at an end, only ONE of the remaining numbers touches B directly, the one right beside it. That number must share a prime with B. From the passage, $B = \{Q, R\}$, so only C, which has Q, or E, which has R, can sit right beside B. We test both choices one at a time and count how many complete, valid rows each choice leads to.

Step 3: Detailed Explanation:
First, place C beside B: row so far is B, C. Now C only has one prime left to offer besides Q, since $C = \{Q, S\}$ and Q is already used against B, so the next number must carry S. Only D carries S, since $D = \{P, S\}$, so the row continues B, C, D.
From D, we can move to A or E, since both carry P, same as D. Trying B, C, D, A: the last spot needs a number sharing a prime with A and it must be E; since A and E both carry P, this works, giving the row B, C, D, A, E.
Trying B, C, D, E instead: the last spot must go to A, and E and A both carry P, so this also works, giving B, C, D, E, A.
So starting with C beside B gives 2 complete valid rows.
Now try E beside B instead: row so far is B, E. Since $E = \{P, R\}$ and R is already used against B, the next number must carry P. Both A and D carry P, so we must check both.
Trying B, E, A: A can lead to D, since they share P, leaving C for the last spot; but C and D share S, so B, E, A, D, C works, a valid row.
Trying B, E, D instead: D can lead to A or C. If it leads to A, the last number is C, but A and C share no prime at all, so this fails. If it leads to C, the last number is A, but C and A again share no prime, so this fails too. So B, E, D gives no valid completion.
So starting with E beside B gives only 1 complete valid row.

Step 4: Final Answer:
Adding the two branches: 2 valid rows starting with C beside B, plus 1 valid row starting with E beside B, gives 3 valid arrangements in total for A, C, D and E.
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