A college admissions committee will grant a certain number

Yes, 18,900 is the correct answer.

This question was also discussed at: college-adminssion-92603.html?hilit=college%20admissions#p713030

Below is my solution from there:

x= number of 10,000 scholarships
y= number of 5,000 scholarships
z= number of 1,000 scholarships
n= total number of scholarships granted = x+y+z = number of students who will receive scholarship (as no student can receive more than one scholarship)

# different ways the committee can dole out the scholarships among the pool of 10 applicants is \(C^n_{10}*\frac{(x+y+z)!}{x!y!z!}=C^n_{10}*\frac{n!}{x!y!z!}\).

\(C^n_{10}\) - choosing \(n\) students who will be granted the scholarship;
\(\frac{n!}{x!y!z!}\) - # of ways we can distribute \(x\), \(y\) and \(z\) scholarships among \(n\) students.

So we need to know the values of \(n\), \(x\), \(y\) and \(z\) to answer the question.

(1) \(n=6\). Don't know \(x\), \(y\) and \(z\). Not sufficient.
(2) \(x=y=z\). Don't have the exact values of \(x\), \(y\) and \(z\) . Not sufficient.

(1)+(2) \(n=6\) and \(x=y=z\). As \(n=6=x+y+z\) --> \(x=y=z=2\) --> # of ways \(C^n_{10}*\frac{n!}{x!y!z!}=C^6_{10}*\frac{6!}{2!2!2!}\). Sufficient.

And again:
\(C^6_{10}\) - choosing \(6\) students who will be granted the scholarship;
\(\frac{6!}{2!2!2!}\) - distributing \(xxyyzz\) (two of each) among 6 students, which is the # of permuatrions of 6 letters \(xxyyzz\).

Answer: C.

Hope it's clear.