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Question Number 184030 by Mastermind last updated on 02/Jan/23

$$\mathrm{How}\mathrm{many}\mathrm{words}\mathrm{can}\mathrm{be}\mathrm{made}$$$$\mathrm{from}5\mathrm{letters}\mathrm{if}$$$$\left(\mathrm{a}\right)\mathrm{all}\mathrm{letters}\mathrm{are}\mathrm{different}$$$$\left(\mathrm{b}\right)2\mathrm{letters}\mathrm{are}\mathrm{identical}$$$$\left(\mathrm{c}\right)\mathrm{all}\mathrm{letters}\mathrm{are}\mathrm{different}\mathrm{but}2$$$$\mathrm{partucular}\mathrm{letters}\mathrm{cannot}\mathrm{be}$$$$\mathrm{adjacent}.$$$$$$$$$$$$\mathrm{M}.\mathrm{m}$$

Answered by mr W last updated on 02/Jan/23

$$\left(a\right)$$$$ABCDE$$$$\Rightarrow 5!=120wordscanbeformed.$$$$\left(b\right)$$$$ABCXX$$$$\Rightarrow \frac{5!}{2!}=60wordscanbeformed.$$$$\left(c\right)$$$$ACBDE$$$$\Rightarrow 5!-4!\times 2!=72wordscanbeformed.$$$$or$$$$\Rightarrow 6\times 2!\times 3!=72$$

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