Question

If $^{n}C_{4}=1365$ then the value of n is equal to

• A. 13
• B. 14
• C. 15
• D. 12
• E. 16

Hint:

Math Tip: Never expand polynomials to the 4th degree (like $n^4 - \dots = 32760$) to solve equations of this type. Always prime factorize the constant and group the factors into consecutive integers.

Answer 1

The Correct Option is C

Concept:
Combinatorics - Combinations Formula.
The formula for combinations is $^{n}C_{r} = \frac{n!}{r!(n-r)!}$, which simplifies to $r$ descending terms of $n$ divided by $r!$.
Step 1: Write out the combination formula for $r=4$.
Expand $^{n}C_{4}$ using the descending factorial method: $$ ^{n}C_{4} = \frac{n(n-1)(n-2)(n-3)}{4 \times 3 \times 2 \times 1} $$ $$ \frac{n(n-1)(n-2)(n-3)}{24} = 1365 $$
Step 2: Isolate the $n$ terms.
Multiply both sides of the equation by 24 to isolate the numerator: $$ n(n-1)(n-2)(n-3) = 1365 \times 24 $$
Step 3: Prime factorize the constant value.
Instead of multiplying to get a huge number (32760), break 1365 into smaller factors: $$ 1365 = 5 \times 273 $$ $$ = 5 \times 3 \times 91 $$ $$ = 5 \times 3 \times 7 \times 13 $$ So, the full expression is: $$ n(n-1)(n-2)(n-3) = (5 \times 3 \times 7 \times 13) \times (8 \times 3) $$
Step 4: Group factors into four consecutive integers.
We need four consecutive integers on the right side to match $n(n-1)(n-2)(n-3)$.
Let's rearrange the factors: $13, 7, 5, 3, 8, 3$.

• One number is $13$.
• We can make $14$ from $7 \times 2$ (borrowing a 2 from the 8, leaving 4).
• We can make $15$ from $5 \times 3$.
• What's left? We have a 4 and a 3 left over, so $4 \times 3 = 12$.

Thus, the right side groups perfectly into: $$ 15 \times 14 \times 13 \times 12 $$
Step 5: Compare and solve for n.
Equate the two sequences of consecutive descending integers: $$ n(n-1)(n-2)(n-3) = 15 \times 14 \times 13 \times 12 $$ By direct comparison of the largest terms: $$ n = 15 $$