Question

EVALUATE EACH OF THE FOLLOWING. LEAVE ANSWER IN REDUCED FRACTION FORM. DO NOT LEAVE ANSWER IN DECIMAL FORM.A) _6C6_8C0 = __________B) 10C3 = __________C) _9P5_10C4 = __________D) _6C2_6P2 = __________E) _7P0_7C0 = _________

Answer 1

Solution A) = 1

Analysis:

$\frac{_6C_6}{_8C_0}=\frac{\frac{6!}{6!(6-6)!}}{\frac{8!}{0!(8-0)!}}$

We know for rules of combinations 0!=1

$\frac{_{6}C_{6}}{_{8}C_{0}}=\frac{\frac{6!}{6!(6-6)!}}{\frac{8!}{0!(8-0)!}}=\frac{\frac{720}{720\ast1}}{\frac{40320}{1\ast40320}}=\frac{1}{1}=1$

Solution B): 120

Analysis

$_{10}C_3=\frac{10!}{3!(10-3)!}=\frac{3628800}{6\ast5040}=\frac{3628800}{30240}=120$

Solution C)=72

Analysis

$\frac{_9P_5}{_{10}C_4}=\frac{\frac{9!}{(9-5)!}}{\frac{10!}{4!(10-4)!}}=\frac{\frac{362880}{24}}{\frac{3628800}{17280}}=\frac{15120}{210}=72$

Solution D)=1/2

Analysis:

$\frac{_6C_2}{_6P_2}=\frac{\frac{6!}{2!(6-2)!}}{\frac{6!}{(6-2)!}}=\frac{\frac{720}{48}}{\frac{720}{24}}=\frac{15}{30}=\frac{1}{2}$

Solution E)=1

Analysis

$\frac{_7P_0}{_7C_0}=\frac{\frac{7!}{(7-0)!}}{\frac{7!}{0!(7-0)!}}=1$