Permutations and Combinations JEE 2025: Prime Products

What is the answer to the prime-product counting question?

The supplied counting method gives 5120 ordered pairs, not its stated final answer of 1124. The permutations and combinations JEE 2025 record needs attribution checks: its year field says 2025, but its date field says 2023-01-24. Neither the year, date nor shift is authenticated until that conflict is resolved.

Let the prime set contain the first ten primes and the product set contain all products formed from distinct members:

S = {p_{1}, p_{2}, \dots, p_{10}}, P = {\prod_{p \in U} p : \emptyset \neq U \subseteq S}, A = S \cup P .

Count ordered pairs whose first coordinate is one of those primes and divides the second coordinate:

x \in S, y \in A, x ∣ y .

A prime cannot occur twice within one product, but may occur in many different products. The record names option C without providing option texts or values; do not reconstruct them.

Why does each product correspond to one non-empty subset?

Each allowed product corresponds to exactly one non-empty subset of the ten primes. Select the primes that occur as factors, then multiply them. Unique prime factorisation guarantees that different subsets cannot give the same integer: changing the selection changes at least one prime factor.

Each prime has two possibilities, included once or excluded. The multiplication principle gives:

\underset{10 choices}{\underset{⏟}{2 \times 2 \times \dots \times 2}} = 2^{10} .

The empty subset has product 1, which the supplied solution excludes. Removing it leaves: $| P | = 2^{10} - 1 = 1023 .$

Under that solution’s convention, a one-prime product is allowed. Singleton subsets already produce the original primes, so taking the union does not add them again:

S \subseteq P, A = S \cup P = P .

The consistent set sizes are:

| S | = 10, | P | = | A | = 1023, | P ⧵ S | = 1013 .

The source’s assignment of 1023 to products outside the prime set is inconsistent. That number counts all non-empty products, including the ten singleton products.

Why does fixing one prime give 512 ordered pairs?

Fixing the first coordinate forces one prime into the second coordinate and leaves nine independent yes-or-no choices. Each fixed prime therefore contributes 512 pairs. Count these choices rather than listing products: unique prime factorisation ensures that each selection gives a different product.

Write the fixed first coordinate as: $x = p_{i} .$

For divisibility, that prime must occur in the product. Every valid second coordinate has the form:

y = p_{i} \prod_{p \in T} p, T \subseteq S ⧵ {p_{i}} .

Each of the remaining nine primes is either present or absent. By the multiplication principle:

# {y : p_{i} ∣ y} = 2^{9} = 512 .

The optional subset may be empty. Its product is 1, so the equal-coordinate pair is already included:

T = \emptyset ⟹ y = p_{i} ⟹ (x, y) = (p_{i}, p_{i}) .

There are ten choices for the first coordinate. Multiplying gives the total:

10 \times 512 = 5120 .

The same second coordinate may appear with different first coordinates because we count ordered pairs, not just distinct products. These are both valid and distinct: $(2, 6) \neq (3, 6) .$

Does the supplied answer of 1124 follow from the solution?

The supplied final answer of 1124 does not follow: its fixed-prime method yields 5120. Adding ten singleton pairs again would give: $5120 + 10 = 5130,$

This is not the source’s stated 1130. The addition also double-counts pairs already included.

The reference to “manual checking” supplies no derivation of 1124. Do not invent one or label 5120 as option C without the options.

Before publication: obtain the original paper, complete options and authenticated answer key before presenting this as a verified PYQ solution. Check the official NTA site, jeemain.nta.nic.in, for source material. The instruction to retain 1124 cannot be satisfied by the supplied mathematics.

Why is adding the pair with equal coordinates a mistake?

The faulty argument counts 512 products for each fixed prime, then adds one more for the prime itself. But excluding all nine optional primes already produces equal first and second coordinates. Adding that case again counts the same pair twice.

The faulty calculation gives: $10 \times (512 + 1) = 5130 .$

Its exact overlap is:

T = \emptyset ⟹ y = x .

Before adding a “special case”, ask: “Which choice in my existing count already produces this case?” Here, the empty optional subset produces it, so nothing should be added.

The value 5130 is an illustrative erroneous result, not a verified supplied option. Wrong-option attribution remains blocked until the missing options are obtained. There is no supplied evidence about which answer students most often choose or find tempting.

How can I test the same counting principle with three related questions?

Change one condition at a time and identify which choices remain free. These are original practice questions, not additional verified PYQs. Each uses non-empty products of distinct primes, so repeated prime factors remain forbidden.

How many pairs are possible with just four primes?

There are 32 ordered pairs. Use this prime set and let the product set contain every non-empty distinct-prime product:

S = {2, 3, 5, 7}, A = {\prod_{p \in U} p : \emptyset \neq U \subseteq S} .

Count the ordered pairs satisfying:

x \in S, y \in A, x ∣ y .

Fix the first coordinate, then choose freely among the other three primes. Repeat for each possible first coordinate:

4 \times 2^{3} = 32 .

Only the number of optional primes changes, from nine to three. The empty optional subset is still allowed.

How many ten-prime products are divisible by both 2 and 3?

There are 256 such products in the original ten-prime setup. Force both factors into the product, then independently include or exclude each of the remaining eight primes:

2^{8} = 256 .

Here, two factors are mandatory. We count products only, so there is no multiplier for choosing a first coordinate.

How many pairs remain if the product needs at least two prime factors?

There are 5110 ordered pairs in the ten-prime setup. For each fixed first coordinate, exclude the empty optional subset because it produces only a singleton product:

10 (2^{9} - 1) = 10 \times 511 = 5110 .

The nine optional primes remain unchanged; only the empty choice is forbidden. Before calculating your next answer, write down the mandatory factors, count the optional primes and decide whether choosing none is allowed.

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Frequently asked questions

How do you get 5120 ordered pairs in the ten-prime problem?

The first coordinate can be any of the ten primes, and the second must be a product of distinct primes containing that chosen prime. For each fixed first coordinate, the other nine primes can independently be included or excluded, giving 2^9 = 512 products. The total is therefore 10 × 512 = 5120 ordered pairs.

Why are there 1023 products of ten distinct primes?

Each non-empty subset of the ten primes gives one product, and unique prime factorisation makes these products distinct. There are 2^10 − 1 = 1023 non-empty subsets. This includes the ten singleton products, so only 1013 products lie outside the original prime set.

Why should we not add the pairs where x equals y separately?

Once the first-coordinate prime is fixed, excluding all nine optional primes gives y = x. That choice is already included in the 512 products counted for each prime. Adding the ten equal-coordinate pairs again double-counts them and incorrectly gives 5130.

Is 5120 the verified official answer to this JEE 2025 question?

No official answer is verified in the supplied record; 5120 is the result of its stated counting method. The record gives 1124 without a supporting derivation, names option C without option values, and has conflicting year and date fields. The original paper, complete options and authenticated answer key are needed before treating it as a verified PYQ.