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 21404 by Tinkutara last updated on 23/Sep/17

$$\mathrm{Prove}\mathrm{that}\left(6n\right)!\mathrm{is}\mathrm{divisible}\mathrm{by}{2}^{2n}.3^n.$$

Answered by dioph last updated on 23/Sep/17

$$\mathrm{if}n>1,\mathrm{there}\mathrm{are}\mathrm{at}\mathrm{least}3n\mathrm{even}$$$$\mathrm{numbers}\mathrm{and}2n\mathrm{multiples}\mathrm{of}3\mathrm{in}$$$$[1,6n],\mathrm{so}\mathrm{that}\left(6n\right)!\mathrm{is}\mathrm{divisible}\mathrm{by}$$$$2^{3n}.3^{2n}\mathrm{and},\mathrm{in}\mathrm{particular},\mathrm{by}{2}^{2n}.3^n$$

Commented by Tinkutara last updated on 23/Sep/17

$$\mathrm{Can}\mathrm{you}\mathrm{explain}{1}^{\mathrm{st}}\mathrm{and}{2}^{\mathrm{nd}}\mathrm{lines}?$$

Commented by dioph last updated on 23/Sep/17

$$\mathrm{from}\mathrm{every}2\mathrm{consecutive}\mathrm{numbers}$$$$1\mathrm{is}\mathrm{even}\mathrm{and}\mathrm{for}\mathrm{every}3\mathrm{consecutive}$$$$\mathrm{numbers}1\mathrm{is}\mathrm{a}\mathrm{multiple}\mathrm{of}3$$

Commented by Tinkutara last updated on 23/Sep/17

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{Sir}!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com