• June 3rd 2010, 02:04 PM
cantcount
1)how many numbers between 2500-6500
using 1,2,3,4,5,6,7 without any repitition

5P1 3P1 7P1 7P1=735

2)Jack needs to answer 5 out of 8 questions find the number of combinations of answers if he can answer any 5 questions

8C5=56

b)if he has to answer 3 out of 4 in paper A and 2 out of 4 in paper B

4C3 4C2= 24

3)6 members must be chosen from 7 boys and 5 girls
a) if there are no restrictions
12C6=924

b)there are not more than 2 girls
5C2 7C4= 350

c) there are 4 girls
5C4 7C2=105

could anyone verify if i am right or wrong.....and please tell me where i went wrong....please
• June 3rd 2010, 02:52 PM
Plato
Quote:

Originally Posted by cantcount
1)how many numbers between 2500-6500
using 1,2,3,4,5,6,7 without any repitition
i answered: 5P1 3P1 7P1 7P1=735

This problem is more complicated than you have considered.
If the number should begin with 2 then the second digit must be 5, 6, or 7.
The last two digits can be $_5P_2=20$. That gives 60 numbers begin with 2.
If the first digit is 6, then the second digit can be 2, 3, or 4. SO?
If the first digit is 3, 4, or 5, then the next three can be $_6P_3=120$. SO?
• June 3rd 2010, 06:12 PM
Soroban
Hello, cantcount!

Quote:

3) 6 members must be chosen from 7 boys and 5 girls

. . b) There are not more than 2 girls

"Not more than 2 girls" means: .0 girls, 1 girl, or 2 girls.

. . $\begin{array}{ccccccc}\text{0 girls, 6 boys:} & (_5C_0)(_7C_6) &=& 1\cdot 7 &=& \;\;7 \\ \text{1 girl, }\text{ 5 boys:} & (_5C_1)(_7C_5) &=&5\cdot21 &=& 105 \\ \text{2 girls, 4 boys:} & (_5C_2)(_7C_4) &=& 10\cdot35 &=& 350 \\ \hline & & & \text{Total:} && 462 \end{array}$