a-word-is-formed-with-3-vowels-and-3-consonants-without-repetition-the-probability-the-formation-of-words-begining-the-letter-z-is-

Question Number 82111 by john santu last updated on 18/Feb/20

$${a}\:{word}\:{is}\:{formed}\:{with}\:\mathrm{3}\:{vowels} \\ $$$${and}\:\mathrm{3}\:{consonants}\:{without}\: \\ $$$${repetition}\:.\:{the}\:{probability}\:{the} \\ $$$${formation}\:{of}\:{words}\:{begining}\:{the} \\ $$$${letter}\:{z}\:{is}? \\ $$

Commented by john santu last updated on 18/Feb/20

i′ got answer p = (C_2 ^(20) /(6×C_3 ^( 21) ))

$${i}'\:{got}\:{answer}\:{p}\:=\:\frac{{C}_{\mathrm{2}} ^{\mathrm{20}} }{\mathrm{6}×{C}_{\mathrm{3}} ^{\:\mathrm{21}} } \\ $$$$ \\ $$

Commented by mr W last updated on 18/Feb/20

p=((C_2 ^(20) C_3 ^5 5!)/(C_3 ^(21) C_3 ^5 6!))=(C_2 ^(20) /(6C_3 ^(21) ))=(1/(42))

$${p}=\frac{{C}_{\mathrm{2}} ^{\mathrm{20}} {C}_{\mathrm{3}} ^{\mathrm{5}} \mathrm{5}!}{{C}_{\mathrm{3}} ^{\mathrm{21}} {C}_{\mathrm{3}} ^{\mathrm{5}} \mathrm{6}!}=\frac{{C}_{\mathrm{2}} ^{\mathrm{20}} }{\mathrm{6}{C}_{\mathrm{3}} ^{\mathrm{21}} }=\frac{\mathrm{1}}{\mathrm{42}} \\ $$

Commented by jagoll last updated on 18/Feb/20

n(S) = 6! C _3^(21) .C_3 ^5 n(A) = 5! C_2 ^(20) .C_3 ^5 p(A) = ((5! C_2 ^(21) .C_3 ^5 )/(6! C_3 ^(21) .C_3 ^5 )) = (C_2 ^(21) /(6.C_3 ^(21) ))

$${n}\left({S}\right)\:=\:\mathrm{6}!\:{C}\:_{\mathrm{3}} ^{\mathrm{21}} .{C}_{\mathrm{3}} ^{\mathrm{5}} \\ $$$${n}\left({A}\right)\:=\:\mathrm{5}!\:{C}_{\mathrm{2}} ^{\mathrm{20}} .{C}_{\mathrm{3}} ^{\mathrm{5}} \\ $$$${p}\left({A}\right)\:=\:\frac{\mathrm{5}!\:{C}_{\mathrm{2}} ^{\mathrm{21}} .{C}_{\mathrm{3}} ^{\mathrm{5}} }{\mathrm{6}!\:{C}_{\mathrm{3}} ^{\mathrm{21}} .{C}_{\mathrm{3}} ^{\mathrm{5}} }\:=\:\frac{{C}_{\mathrm{2}} ^{\mathrm{21}} }{\mathrm{6}.{C}_{\mathrm{3}} ^{\mathrm{21}} } \\ $$

Commented by mr W last updated on 18/Feb/20