If the club decides to send 5 members to the convention, how many different groups of 5 are possible? there are 10 members

1 answer:

7 0

For the 1st member, you have 10 possible choices.

For the 2nd, you have 9 remaining choices.

For the 3rd: 8.

So all possible combinations would be:
10*9*8*7*6 = 10! - 5! = 30,240 combinations.

You might be interested in

Which terms in the expression 9d2+15d3+11c2+7 have 14c2 as a like term?

KatRina [158]

That would be the term 11c2 because like 14c2, 11c2 has a constant, a variable and raised to the power of 2.

3 0

2 years ago

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Rom4ik [11]

Answer:

Step-by-step explanation:

100- 51 = 49

7 0

3 years ago

Complementary of supplementary? Find the value of x

prohojiy [21]

Answer: x=6

The two angles shown in each are complementary because they add up to 90°.

10 & 12 would be supplementary to one another because they would add up to 180°.

Step-by-step explanation:

We know that on both 10 & 12 the angles add up to equal 90° so...

10. 8x+7x=90

15x=90

x=6

12. it's the same in pic as 10

The two angles shown in each are complementary because they add up to 90°.

10 & 12 would be supplementary to one another because they would add up to 180°.

5 0

2 years ago

Help me please ......

Alja [10]

$2\frac{3}{4}*2=\frac{2*4+3}{4}*2=\frac{11}{4}*2=\frac{22}{4}= 5\frac{2}{4}=5\frac{1}{2}$

7 0

3 years ago

5 (RootIndex 3 StartRoot x EndRoot) + 9 (RootIndex 3 StartRoot x EndRoot)

Ratling [72]

Answer:

C. 14 RootIndex 3 StartRoot x EndRoot)

Step-by-step explanation:

To obtain the sum ; the a clear explanation has been attached in the picture below ;

8 0

2 years ago