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Question-122973




Question Number 122973 by bemath last updated on 21/Nov/20
Answered by mr W last updated on 21/Nov/20
12.  ((10+C_2 ^3 ×10)/(10^3 ))=(1/(25))=4%
$$\mathrm{12}. \\ $$$$\frac{\mathrm{10}+{C}_{\mathrm{2}} ^{\mathrm{3}} ×\mathrm{10}}{\mathrm{10}^{\mathrm{3}} }=\frac{\mathrm{1}}{\mathrm{25}}=\mathrm{4\%} \\ $$
Commented by bemath last updated on 21/Nov/20
yes...
$${yes}… \\ $$
Answered by mr W last updated on 21/Nov/20
13.  a)  (C_3 ^(12) −C_1 ^(10) )×3!=1260  b)  (C_1 ^(10) +C_3 ^(10) )×3!=780  c)  C_2 ^(11) ×3!=330  d)  C_2 ^(11) ×2!+C_3 ^(11) ×3!=1430  e)  C_1 ^(10) ×3!=60  f)  (1+2×9)×3!=114
$$\mathrm{13}. \\ $$$$\left.{a}\right) \\ $$$$\left({C}_{\mathrm{3}} ^{\mathrm{12}} −{C}_{\mathrm{1}} ^{\mathrm{10}} \right)×\mathrm{3}!=\mathrm{1260} \\ $$$$\left.{b}\right) \\ $$$$\left({C}_{\mathrm{1}} ^{\mathrm{10}} +{C}_{\mathrm{3}} ^{\mathrm{10}} \right)×\mathrm{3}!=\mathrm{780} \\ $$$$\left.{c}\right) \\ $$$${C}_{\mathrm{2}} ^{\mathrm{11}} ×\mathrm{3}!=\mathrm{330} \\ $$$$\left.{d}\right) \\ $$$${C}_{\mathrm{2}} ^{\mathrm{11}} ×\mathrm{2}!+{C}_{\mathrm{3}} ^{\mathrm{11}} ×\mathrm{3}!=\mathrm{1430} \\ $$$$\left.{e}\right) \\ $$$${C}_{\mathrm{1}} ^{\mathrm{10}} ×\mathrm{3}!=\mathrm{60} \\ $$$$\left.{f}\right) \\ $$$$\left(\mathrm{1}+\mathrm{2}×\mathrm{9}\right)×\mathrm{3}!=\mathrm{114} \\ $$

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