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It is an exercise question (#19) in the book with image showing all details.

My answer is given below, & request vetting. $$\sum_{k=19}^{20} \binom{20}{x}\cdot 3^{(20-k)} $$ $$\binom{20}{19}\cdot 3^1 => 60$$ enter image description here

jiten
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    This is clearly not the way one is supposed to ask questions on MSE. Nontheless. the first formula you wrote is correct. – b00n heT Sep 03 '18 at 06:11

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You almost got it, $(0.9 ) 20 = 18$, it should be

$$\sum_{k=\color{red}{18}}^{20} \binom{20}{x}\cdot 3^{(20-k)} $$

Siong Thye Goh
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  • Thanks for pointing the error. – jiten Sep 03 '18 at 06:19
  • Please help with my post of today at : https://math.stackexchange.com/q/2903732/424260. It is regarding detailed analysis of the 'misapplication of the product principle'. – jiten Sep 03 '18 at 10:19
  • Please help with my last comment, as no response is there till now. – jiten Sep 03 '18 at 10:51
  • Is my earlier comment's post (for which I asked help) is obscure or incomprehensible; so that I need to restate it. I need help in that post, to clear up my understanding, & have failed to elicit any response so far. Please tell me what it lacks. – jiten Sep 03 '18 at 12:48
  • Please tell something, as have lost hope on my solving it on my own. Also, that would be helpful to me as a guideline for future posts. – jiten Sep 03 '18 at 13:04
  • Can we chat on the issues concerning my stated post at : https://math.stackexchange.com/q/2903732/424260 – jiten Sep 03 '18 at 13:13
  • I found it. Please see the update on the post & vet. – jiten Sep 03 '18 at 13:37
  • Am I committing some sort of behavioral error, that there is no response from you? – jiten Sep 03 '18 at 14:26
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    I am still outside of my house. – Siong Thye Goh Sep 03 '18 at 14:30