+0

# help with some word problems

-1
210
6

How many five-digit numbers have distinct digits which are decreasing from left to right? (For example, 96531 is such a number.)

A committee of 4 is to be chosen from a group of students. If the number of students in the group increases by 1, the number of different committees doubles. How many students are in the group?

Thanks!

May 4, 2019
edited by Melody  May 5, 2019

#1
+2552
+1

I don't have time so I'm going to answer the last one

If you are to choose 4 out of x students it will be $$\frac{x!}{4!(x-4)!}=y$$

Then if the number of students in the group increases by 1, the committees double.

So $$\frac{(x+1)!}{4!((x+1)-4)!}=2y$$

So we have two equations with 2 variables.

You can try and solve now, (try substituting)

May 4, 2019
#4
0

How do u subsitute?

Guest May 5, 2019
#5
0

nvm, i got it

Guest May 5, 2019
#2
+1

How many five-digit numbers have distinct digits which are decreasing from left to right? (For example, 96531 is such a number.)

1- ABCDE. I will give this a shot.

2 - For A, you have 9 choices.

3 - For B, you still have 9 choices out of 10, because we are excluding A

4 - For C, you have 8 choices because we are excluding A and B

5 - For D, you have 7 choices because we are excluding A, B, and C.

6 - For E, you have 6 choices because we are excluding A, B, C and D.

7 - Then we have: 9 x 9 x 8 x 7 x 6 =27,216 - numbers.

8 - The biggest such number would be: 98765.

9 - The smallest such number would be: 43210

10 - The difference:98765 - 43210 =55,555 from which 27,216 would be descending numbers.

Note: somebody should check this. Thanks.

May 4, 2019
#3
0

No, thats not right

Guest May 5, 2019
#6
0

$${10\choose 5}=\dfrac{10\cdot9\cdot8\cdot7\cdot6}{5\cdot4\cdot3\cdot2\cdot1}=9\cdot 4\cdot 7=\boxed{252}$$