n+1CrnCr=5535 \frac{{n+1}C_r}{nC_r} = \frac{55}{35} nCrn+1Cr=3555
(n+1)!(r+1)!(n−r)!÷n!r!(n−r)!=117 \frac{(n+1)!}{(r+1)!(n-r)!} \div \frac{n!}{r!(n-r)!} = \frac{11}{7} (r+1)!(n−r)!(n+1)!÷r!(n−r)!n!=711
n+1r+1=117 \frac{n+1}{r+1} = \frac{11}{7} r+1n+1=711
7n=4+11r 7n = 4 + 11r 7n=4+11r
nCrn−1Cr−1=3521 \frac{nC_r}{n-1C_{r-1}} = \frac{35}{21} n−1Cr−1nCr=2135
n!r!(n−r)!÷(n−1)!(r−1)!(n−r)!=53 \frac{n!}{r!(n-r)!} \div \frac{(n-1)!}{(r-1)!(n-r)!} = \frac{5}{3} r!(n−r)!n!÷(r−1)!(n−r)!(n−1)!=35
nr=53 \frac{n}{r} = \frac{5}{3} rn=35
3n=5r 3n = 5r 3n=5r
Solving for r=6 r = 6 r=6 and n=10 n = 10 n=10:
2n+5r=50 2n + 5r = 50 2n+5r=50
Find the equivalent capacitance between A and B, where C=16 μF C = 16 \, \mu F C=16μF.