Solve the following problems:

To form a task force consisting of one member of X and one member of Y, you can use the concept of combinations. Since committee X has 4 members and committee Y has 5 members, you have choices for selecting one member from each committee.

The number of different task forces possible is the product of the number of ways you can choose one member from each committee.

Number of ways to choose one member from X = C(4, 1) = 4 (choose 1 from 4) Number of ways to choose one member from Y = C(5, 1) = 5 (choose 1 from 5)

Total number of different task forces = 4 (from X) * 5 (from Y) = 20

So, there are 20 different task forces possible.

$Total value of initial configeration = 13 hamper \times \$12 = \$156$

$Again total value of new configeration = 12 hamper \times \$13=\$156$

$\therefore$The value of gift certificates will not be changed.

$$\begin{gathered} Now Susan started with 13 hampers. \\ Susan has 13 piece of \$3 worth gift \end{gathered}$$

$$\begin{gathered} \therefore Susan has 13 piece of \$4 worth gift \\ \therefore Susan has 13 piece of \$5 worth gift \\ \therefore Possible set \{5, 5, 3\} \{4, 4, 5\} , (3, 3, 3, 4) in which \$5 contains in first two of them and the total \end{gathered}$$

number is 13.

Now, let Susan creates

x hampers of {3, 3, 3, 4}

y hampers of {4,4,5}

And z hampers of {5, 5, 3}

Number of $3 gift will be, 3x + z =13.....(i)

$\therefore Number of \$4 gift will be, x + 2y =13.....\left(ii\right)$

$$\begin{gathered} \therefore Number of \$5 gift will be, y + 2z = 13 \\ Now, multiplying equation \left(iii\right) by 2 and subtracting \left(ii\right) from \left(iii\right) we get , \end{gathered}$$

$$\begin{gathered} \frac{2y+4z=26 \\ x+2y=13 \\ - - - }{4z-x=13...............\left(iv\right)} \end{gathered}$$

$$\begin{gathered} Again, multiplying equation \left(i\right) by 4 and subtracting equation \left(iv\right) from \left(i\right) we get \\ \frac{12x+4z=52 \\ 4z-x=13}{13x=39} \\ \therefore x=3 \\ Here, x is the number of gift hampers without \$5 certificates. \\ \therefore 3 gift hampers did not contain \$5 gift certificates. \end{gathered}$$