There are five salesmen abcd and e who can talk in different languages a company wants to make

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We can choose 3 salesmen out of 5 in "C^3_5=\\frac{5!}{3!\\cdot2!}=10" ways. So we have 10 possible combinations of 3 elements out of 5, these combinations define our sample space.

(i) If one of them is A, then we can choose 2 more salesmen out of 4 remaining (A is already chosen). We can do this in "C^2_4=\\frac{4!}{2!\\cdot2!}=6" ways. Thus, the probability, one of the salesmen is A equals 6/10=0.6.

(ii) If A is not selected, 3 salesmen out of 4 is chosen (A is now removed). It can be done in "C^3_4=\\frac{4!}{3!\\cdot1!}=4" ways. The probability of this is 4/10=0.4.

(iii) If A is chosen, we should choose 2 more salesmen out of 4. But we can't choose B, so now we should choose 2 salesmen out of 3. There are "C^2_3=\\frac{3!}{2!\\cdot1!}=3" ways to do this. The same situation will take place if we choose B without A. Thus, we have 3+3=6 ways to choose A or B (not both). So the probability of this situation is 6/10=0.6.

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Related There are 5 speakers A, B, C, D and E. The number of ways in which A will speak always before B isa)24b)c)d)none of theseCorrect answer is option 'A'. Can you explain this answer?

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Is the arrangement CDABE I included in it as it satisfy the given question
explain me??

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Answer will be 6 damn first you have to arrange the members C D E in five places in 5p3 ways and you will have two places left over and fill it in the order A and B such that A will deliver his speech before B

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Related There are 5 speakers A, B, C, D and E. The number of ways in which A will speak always before B isa)24b)c)d)none of theseCorrect answer is option 'A'. Can you explain this answer?

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Related There are 5 speakers A, B, C, D and E. The number of ways in which A will speak always before B isa)24b)c)d)none of theseCorrect answer is option 'A'. Can you explain this answer?

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