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Maths Question 15 – AP-EAMCET 2026

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If n is an integer between 0 and 31, then the minimum value of n!(31−n)! is

The product n!(N−n)! tends to be minimal when n and N−n are as close as possible.

Step 1: Define the Function and Domain✦ Active

Let the given expression be represented by the function f(n)=n!(31−n)!. We are looking for the minimum value of this function. The problem states that n is an integer between 0 and 31. This implies that n∈{1,2,…,30}.

💡 Teacher's Secret Hint

Remember that 'between a and b' usually excludes a and b unless specified otherwise. In this case, n=0 or n=31 would result in 31!, which is not the minimum.

Step 2: Analyze Function Behavior using Ratios○ Expand

To find the minimum value, we can analyze how the function changes as n increases by looking at the ratio of consecutive terms, f(n+1)f(n).

f(n+1)f(n)=(n+1)!(31−(n+1))!n!(31−n)!

Simplify the expression using the property k!=k×(k−1)!:

f(n+1)f(n)=(n+1)×n!×(30−n)!n!×(31−n)×(30−n)!=n+131−n
💡 Teacher's Secret Hint

This ratio method is a common technique for finding extrema of functions involving factorials or binomial coefficients.

Step 3: Determine the Minimum Point○ Expand

The function f(n) is decreasing when f(n+1)f(n)<1 and increasing when f(n+1)f(n)>1. The minimum will occur where the function transitions from decreasing to increasing.

1. If f(n+1)<f(n) (decreasing): n+131−n<1⟹n+1<31−n⟹2n<30⟹n<15. So, for n=1,2,…,14, the function is decreasing.

2. If f(n+1)>f(n) (increasing): n+131−n>1⟹n+1>31−n⟹2n>30⟹n>15. So, for n=16,17,…,30, the function is increasing.

This means the function decreases up to n=15 and then starts increasing from n=16. Therefore, the minimum value occurs at n=15 or n=16. Let's check these values:

💡 Teacher's Secret Hint

The minimum value typically occurs when the terms involved in the product are as close to each other as possible. Here, n and 31−n are closest when n≈(31)/2=15.5, leading to integer values of n=15 or n=16.

Step 4: Calculate the Minimum Value○ Expand

Substitute n=15 into the function:

f(15)=15!(31−15)!=15!16!

Substitute n=16 into the function:

f(16)=16!(31−16)!=16!15!

Both values are identical: 15!16!. This is the minimum value of the expression within the given range.

💡 Teacher's Secret Hint

Always check both integer values closest to the theoretical minimum point, especially when the theoretical minimum is not an integer.

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