There are 6 speakers at a conference: Ms. Sally, Ms. Elaine, Ms. Boo-Koo, Mr. Adam, Mr. Jones, and Mr. Tall. In how many ways can we order the speakers so that Mr. Adam doesn't speak first?

Question

Mathematics

Javon Rollins

9 March, 23:55

There are 6 speakers at a conference: Ms. Sally, Ms. Elaine, Ms. Boo-Koo, Mr. Adam, Mr. Jones, and Mr. Tall. In how many ways can we order the speakers so that Mr. Adam doesn't speak first?

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Silas Kennedy · Answer 1

Problems such as this are called counting problems. They often ask "in how many way" something can occur. Here we have 6 speakers and need to arrange them in order. We can use the counting principle which tells us that the number of ways can be obtained by considering how many speakers could be chosen for each position (first, second, third, ... sixth) and multiplying these.

Even though there are 6 speakers, Mr. Adam cannot go first so there are 5 choices for who goes first. That leaves 5 speakers who can go second. As we already picked two people, there are 4 speakers who can go third and 3 who can go fourth. That leaves 2 that can go fifth and 1 that is left for the last slot.

The total number of ways is given by (5) (5) (4) (3) (2) (1) = 600