Out of all possible 8 digit numbers formed us | vedclass

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Question

Total number of numbers $ = {\,^7}{{\rm{C}}_2} \cdot \frac{{6!}}{{2!2!}} = {\rm{a}}$

for odd numbers:

Numbers with unit's digit $3:$

$^{6

Vedclass · Accepted Answer

$\frac{5}{14}$

Vedclass · Answer

Total number of numbers $ = {\,^7}{{m{C}}_2} \cdot \frac{{6!}}{{2!2!}} = {m{a}}$

for odd numbers:

Numbers with unit's digit $3:$

$^{6} \mathrm{C}_{2} \cdot \frac{5 !}{2 ! 2 !}=\mathrm{b}$

Numbers with unit's digit $1:$

$^{6} \mathrm{C}_{2} \cdot \frac{5 !}{2 !}=\mathrm{c}$

$	herefore $ Required probability $=\frac{b+c}{a}=\frac{5}{14}$

Out of all possible $8$ digit numbers formed using all the digits $0,0,1,1,2,3,4,4$ a number is randomly selected. Probability that the selected number is odd, is-

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