Question

A man desires to throw a party for some of his friends. In how many ways can he select 8 friends from a group of 11 friends, if the two of his friends(say ’A’ and ’B’) will not attend the party together?

Answer 1

Answer:

A+B= 2 members

11-2=9

remove a friend that do not behave good than the other 8

8

Step-by-step explanation:

i hope it helps mark me brainliest

Answer 2

Concept introduction:

A permutation is an orderly arrangement of the items or numbers. Pairings are a means to choose items or numbers from a collection or set of items without regard for the items' chronological sequence.

Given:

Here it is given that  he select $8$ friends from a group of $11$ friends, if the two of his friends(say ’$A$’ and ’$B$’) will not attend the party together.

To find:

We have to find how many ways he can select his friends.

Solution:

According to the question, Since $A$ and $B$ will not attend together, there is only $10$ friends to choose from:

So the ways are $10C_{8}$.

                           $= > 1287$.

Final answer:

Hence the final answer of the question is $1287$ ways.

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