The value of \(^nC_0 + ^nC_3 + ^nC_6 + ^nC_9\) is \(1 + \frac{n!}{6}[\frac{1}{(n - 3)!}+\frac{1}{120(n - 6)!} + \frac{1}{60480(n - 9)!}]\)
The expression is given as:
\(^nC_0 + ^nC_3 + ^nC_6 + ^nC_9\)
The combination formula is:
\(^nC_r = \frac{n!}{(n - r)!r!}\)
So, we have:
\(\frac{n!}{(n - 0)!0!} + \frac{n!}{(n - 3)!3!}+\frac{n!}{(n - 6)!6!} + \frac{n!}{(n - 9)!9!}\)
Evaluate the factorials
\(\frac{n!}{(n - 0)!} + \frac{n!}{6(n - 3)!}+\frac{n!}{720(n - 6)!} + \frac{n!}{362880(n - 9)!}\)
Evaluate the difference
\(\frac{n!}{n!} + \frac{n!}{6(n - 3)!}+\frac{n!}{720(n - 6)!} + \frac{n!}{362880(n - 9)!}\)
Evaluate the quotient
\(1 + \frac{n!}{6(n - 3)!}+\frac{n!}{720(n - 6)!} + \frac{n!}{362880(n - 9)!}\)
Factor out n!
\(1 + n![\frac{1}{6(n - 3)!}+\frac{1}{720(n - 6)!} + \frac{1}{362880(n - 9)!}]\)
Factor out 1/6
\(1 + \frac{n!}{6}[\frac{1}{(n - 3)!}+\frac{1}{120(n - 6)!} + \frac{1}{60480(n - 9)!}]\)
Hence, the value of \(^nC_0 + ^nC_3 + ^nC_6 + ^nC_9\) is
\(1 + \frac{n!}{6}[\frac{1}{(n - 3)!}+\frac{1}{120(n - 6)!} + \frac{1}{60480(n - 9)!}]\)
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