# How many committees of 5 members are there?

## How many committees of 5 members are there?

5! The variety of methods a selected official member is rarely included = 462 methods. (d)We have to search out the entire variety of methods a selected member is at all times included within the committees. Total variety of methods a selected member is at all times included within the committees= 11C4=11!

252 methods

40320

## How some ways can 8 be organized?

40,320 totally different mixtures

40,320 methods

## How many alternative methods 8 folks can sit?

So for 8 seats there are 8x7x6x5x4x3x2x1 = 8! potential methods to seat 8 folks.

40320 methods

40,320

## How some ways can 8 individuals line up for a photograph?

SOLUTION: How some ways are you able to prepare 8 folks in a row for {a photograph}? (Points : 1) 630 5040

## How some ways can 4 folks be chosen and organized in a straight line if there are 9 folks to select from?

In 12*11*10*9 = 11880 methods. ANSWER Explanation : Any of 12 individuals on the 1-st place. Any of 11 remaining individuals on the 2-nd place. Any of 10 remaining individuals on the 3-rd place.

## How some ways can 4 individuals be organized in a straight line?

A bunch of 4 individuals are standing in a straight line. In what number of alternative ways can these folks be standing on the road? The reply is 24.

## How some ways can 5 folks be chosen and organized in a straight line if there are 12 folks to select from?

Once you’ve got chosen the 5 folks, there are 5!, or 120 alternative ways to line them up.

## How some ways can 10 folks stand in a line?

Assuming they’re multi functional line, we are able to line them up in 10! methods: there are 10 individuals who can stand in place 1, then 9 remaining who can stand in place 2, then 8 remaining who can stand in place 3, and so on. If we add the power to maneuver the ten locations the place folks can stand, then we’ll should multiply 10!

## How some ways can banana be organized?

There are 3×2×1 = 6 methods to rearrange the A’s and a couple of×1 = 2 methods to rearrange the N’s. So, there are: 720/(6×2) = 720/12 = 60 methods to rearrange all of the letters in BANANA.

## How some ways can 9 college students line up for lunch?

Overall, it seems like 9x8x7x6x5x4x3x2x1, or 362,880 methods of arranging the folks.

362,880 methods

## How some ways are you able to order 7 issues?

Answer: There are 7 × 6 × 5 × 4 × 3 × 3 × 1 = 7! = 5040 methods.

## How many occasions can 7 numbers be organized?

=7⋅6⋅5⋅4⋅3⋅2⋅1=5040. This explicit drawback is a permutation. Recall, the distinction between permutations and mixtures is that, with permutations, order issues.

## How some ways can 3 numbers be organized?

There are, you see, 3 x 2 x 1 = 6 potential methods of arranging the three digits. Therefore in that set of 720 potentialities, every distinctive mixture of three digits is represented 6 occasions. So we simply divide by 6.