Please answer all parts to the questions with explanation. Thanks. 4. A committe

Question

Please answer all parts to the questions with explanation. Thanks.

4. A committee of three is to be selected from a group of 20 students, five each from four different schools. How many such committees consist of students all from the same school? 5. In how many ways can five adults and four children line up in a row to get into a movie if the first three in line must be adults? 6. How many five letter "words can be made from the set of letters fa,b,c, d, e, f, g, h) if (a) the "word" must contain the letter g and no letter can be used more than once? (b) the word" must contain exactly two e's, and no other letter can be used more than once? (c) the word" must contain exactly one vowel? 7. How many five element subsets of the set {1,2,3, 4,5,6, 7,8, 9 are there that (a) contain both 3 and 5? (b) contain 3 but not 5? (c) contain exactly two even mumbers? 8. An executive committee of president, vice-president, secretary, and treasurer is to be chosen from a group of four men and five women. In how many ways can this be done if a) the president and secretary must be women? (b) Beth refuses to be secretary? (c) there must be one and only one man on the executive? (d) Andy and Beth must be on the executive?

Explanation / Answer

Solved first two problems in detail, As per the chegg guidelines, please post multiple question to get the remaining answers

Q4) There are four school with five members each

Comittee consisting students from same school = 5C3 (ways, we can choose any 3 students from group of 5 students of same school) = 5!/3!2! = 10 ways

Since there are total of 4 schools, hence number of ways = 4C1(select any school) * 5C3 (ways, we can choose any 3 students from group of 5 students of same school)

=> 4 * 10 = 40 ways

5)

There are 5 adults and 4 children

Necessary condition is that the first three must be adults

There are 9 slots to be filled, first three slots must be filled by adults

Choose any 3 adults = 5C3 ways

Now permute these adults in 3! ways, now we are left with 2 adults and 4 childens, whose position didn't matter, hence we can arrange them in 6! ways

So, Total ways are 5C3 * 3! * 6! = 43200 ways

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