Suppose that s be the set of all ordered 4 tuplets(x,y,z,w) of the positive integers which are the solutions of x+y+z+w=21 .one such ordered tupleof solutions is selected at random from s .then find the probability that x>y

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$$\begin{gathered} \mathrm{The} \mathrm{total} \mathrm{number} \mathrm{of} \mathrm{solution} \mathrm{will} \mathrm{be} {}^{20}\mathrm{c}_3 \\ \mathrm{x} \mathrm{will} \mathrm{be} \max 18 . \\ \mathrm{x} \mathrm{will} \mathrm{be} \mathrm{minimum} 2 \\ \mathrm{if} \mathrm{x}=2 \mathrm{then} \mathrm{y}=1 \\ \mathrm{if} \mathrm{x}=3 \mathrm{then} \mathrm{y}=2 \\ \mathrm{And} \mathrm{so} \mathrm{on} \mathrm{till} \\ \mathrm{x}=18 \mathrm{then} \mathrm{y}=17 \\ \mathrm{So} 1+2+3+4....+17=\frac{17\left(17+1\right)}{2} \mathrm{because} 1+2+3+4...\mathrm{nterms}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2} \\ \mathrm{So} 1+2+3+4...+17=17\times 9 \\ \mathrm{And} \mathrm{now} \mathrm{the} \mathrm{probability} \mathrm{in} \left(\mathrm{x}>\mathrm{y}\right)=\frac{17\times 9}{ {}^{20}\mathrm{c}_3} \\ \frac{17\times 9}{{\displaystyle \frac{20!}{3!\times 17!}}}=\frac{17\times 9\times 3!\times 17!}{20!}=\frac{17\times 9\times 3\times 2\times 1\times 17!}{20\times 19\times 18\times 17!}=\frac{51}{20\times 19} \\ \mathrm{Probability} \mathrm{will} \mathrm{be} \frac{51}{380} \mathrm{Answer} \end{gathered}$$


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