Question and Answers Forum

All Questions      Topic List

Permutation and Combination Questions

Previous in All Question      Next in All Question      

Previous in Permutation and Combination      Next in Permutation and Combination      

Question Number 50724 by vajpaithegrate@gmail.com last updated on 19/Dec/18

$$\mathrm{a}\mathrm{man}\mathrm{goes}\mathrm{in}\mathrm{for}\mathrm{an}\mathrm{examination}\mathrm{in}$$$$\mathrm{which}\mathrm{there}\mathrm{are}4\mathrm{papers}\mathrm{which}\mathrm{maxmum}$$$$\mathrm{of}10\mathrm{marks}\mathrm{for}\mathrm{each}\mathrm{paper}\mathrm{the}\mathrm{no}\mathrm{of}\mathrm{ways}$$$$\mathrm{of}\mathrm{getting}20\mathrm{marks}\mathrm{on}\mathrm{the}\mathrm{whole}\mathrm{is}$$$$\mathrm{ans}:891$$

Answered by ajfour last updated on 19/Dec/18

$$=coeff.of{x}^{20}in{(1+x+x^2+...+x^{10})}^4$$$$=coeff.of{x}^{20}in{(1-x^{11})}^4{(1-x)}^{-4}$$$${=}^{4+20-1}{C}_3-4\left({}^{4+9-1}{C}_3\right)$$$$={}^{23}{C}_3-4\left({}^{12}{C}_3\right)$$$$=\frac{23\times 22\times 21}{6}-4\times \frac{12\times 11\times 10}{6}$$$$=23\times 11\times 7-4\times 2\times 11\times 10$$$$=11(161-80)=11\times 81=891.$$

Commented by vajpaithegrate@gmail.com last updated on 19/Dec/18

$$\mathrm{thank}\mathrm{u}\mathrm{sir}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com