Question Number 40284 by scientist last updated on 18/Jul/18
$${How}\:{many}\:{permutation}\:{can}\:{be}\:{made}\:{using}\:{the}\:{word} \\ $$$${CANADA}\:{each}\:{of}\:{the}\:{six}\:{letter}\:{being}\:{used}\:{mixed} \\ $$$${in}\:{each}\:{permutation}?\:\:{In}\:{how}\:{many}\:{of}\:{these} \\ $$$${permutation}\:{will}\:{the}\:{three}\:{A}'{s}\:{be}\:{together}?\:{In}\: \\ $$$${how}\:{many}\:{will}\:{two}\:{A}'{s}\:{be}\:{together}\:{but}\:{not}\:{the}\:{three}? \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 19/Jul/18
$$\left.{i}\right){CANADA}\:\:\:{A}=\mathrm{3}\:{C}=\mathrm{1}\:{D}=\mathrm{1}\:{N}=\mathrm{1} \\ $$$${permutaion}\:{numbers}=\frac{\mathrm{6}!}{\mathrm{3}!}=\frac{\mathrm{6}×\mathrm{5}×\mathrm{4}×\mathrm{3}!}{\mathrm{3}!}=\mathrm{120} \\ $$$$\left.{ii}\right){let}\:\mathrm{03}\:{numbers}\:{of}\:{A}\:{faviquicked}\:{with}\:{each} \\ $$$${other}\:{so}\:\left({AAA}\right){CDN}\: \\ $$$${permutation}=\mathrm{4}!=\mathrm{24} \\ $$$$\left.{iii}\right)\left({AA}\right){ACDN} \\ $$$${pdrmutation}=\mathrm{3}{C}_{\mathrm{2}} ×\mathrm{5}!=\mathrm{3}×\mathrm{120}=\mathrm{360} \\ $$$${pls}\:{check}… \\ $$$${after}\:{editing}… \\ $$$$\left.{iii}\right){required}\:{answer}\:{is}\:\mathrm{360}−\mathrm{24}=\mathrm{336} \\ $$$${becos}\:\:{out}\:{of}\:{these}\:\mathrm{360}\:{permutation}\:\:\mathrm{24} \\ $$$${permutation}\:{are}\:{included}\:{where}\:\mathrm{03}\:\:{AAA}\:{will} \\ $$$${be}\:{together}…{so}\:{ans}\:{is}\:\mathrm{360}−\mathrm{24}=\mathrm{336} \\ $$
Commented by MJS last updated on 18/Jul/18
$$\left.{iii}\right)\:\mathrm{might}\:\mathrm{be}\:\mathrm{wrong}\:\mathrm{because}\:{AAA}\:\mathrm{is}\:\mathrm{not} \\ $$$$\mathrm{allowed}\:\mathrm{but}\:\mathrm{I}'\mathrm{m}\:\mathrm{not}\:\mathrm{sure}\:\mathrm{in}\:\mathrm{these}\:\mathrm{things} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 19/Jul/18
$${yes}\:{you}\:{are}\:{right}…{i}\:{am}\:{editing}… \\ $$