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Bunuel , KarishmaB , ScottTargetTestPrep Please help with this question. I am not able to figure it out.
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no of digits=D
no of vowels=v
no of consonants=c

=> d(d-1)/9.10=0.1 -0.33
=> v(v-1)/9.10=0.01-0.1
=> d=4,5
=> v=2,3
=> (d+v)_min=6, (d+v)_max=8
=> c_max=4, c_min=2

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­For first condition (to choose two digits ) there are only 2 possible scenarios 5C2/10C2 or 4C2/10C2 which gives us probability b/w 10 & 33% .... the no of possible digits are 4 or 5

this implies that no of alphabets are either 5 or 6 

Using second condition to choose vowels there are again only 2 possible scenarios 2C2/10C2 or 3C2/10C2 which gives us probability between 1 & 10% 

Now if there are (2,3) vowels out of (5,6) alphabets then max consonents: 6-2 =4 & min consonents: 5-3 = 2


 ­
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chetan2u
If element is just 1: \(\frac{1}{10} * \frac{0}{9} = 0\)%
If there are two elements: \(\frac{2}{10} * \frac{1}{9} = \frac{1}{45} \ \ or \ \ \frac{100}{45} \) % = 2.xy%
If there are three elements: \(\frac{3}{10} * \frac{2}{9} = \frac{1}{15} \ \ or \ \ \frac{100}{15}\)% = 6.xy %
If there are four elements: \(\frac{4}{10} * \frac{3}{9} = \frac{2}{15} \ \ or \ \ \frac{200}{15}\)% = 12.xy %
If there are five elements: \(\frac{5}{10} * \frac{4}{9} = \frac{2}{9} \ \ or \ \ \frac{200}{9}\)% = 22.22 %
If there are six elements: \(\frac{6}{10} * \frac{5}{9} = \frac{1}{3} \ \ or \ \ \frac{100}{3}\)% = 33.33 %
chetan2u can you please explain how did you form these equations? Very confused
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Solution:
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