Quiz Level-3

Hint

\(
\begin{aligned}
& \left(\begin{array}{l}
n \\
2
\end{array}\right)=\frac{n(n-1)}{2}=4560 ; n>0 \\
& n^2-n-9120=0 \\
& a=1 ; b=-1 ; c=-9120 \\
& D=b^2-4 a c=1^2-4 \cdot 1 \cdot(-9120)=36481 \\
& D>0 \\
& n_{1,2}=\frac{-b \pm \sqrt{D}}{2 a}=\frac{1 \pm \sqrt{36481}}{2} \\
& n_{1,2}=\frac{1 \pm 191}{2} \\
& n_{1,2}=0.5 \pm 95.5 \\
& n_1=96 \\
& n_2=-95
\end{aligned}
\)
Factored form of the equation:
\(
(n-96)(n+95)=0
\)
\(
n>0 \Rightarrow n=96
\)

Hint

\(
n=2 \cdot 6=12
\)

Hint

If the numbers do not have to be different: there would be \(16^3=4096\) combinations ( 16 for each of the first, second, and third choices).

Hint

As the three numbers need to be distinct: then there are 40 choices for the first number; then 39 choices for the second number (having eliminated one number); then 38 choices for the third number (having eliminated two numbers).
\(
40 \times 39 \times 38=59,280
\)

Hint

Take any path and represent it as series of moves, either U (up) or R (right). Every unique 7-unit path consists of four R’s and three U’s. The question then becomes: How many ways can we rearrange the letters RRRRUUU:
\(
\frac{7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1}{(3 \cdot 2 \cdot 1)(4 \cdot 3 \cdot 2 \cdot 1)}=35
\)

Hint

Each rectangle requires that we choose two horizontal and two vertical lines. There are \({ }^4 C_2=6\) ways to choose two horizontal lines and \({ }^7 C_2=21\) ways to choose two vertical lines, which gives us \(21 \cdot 6=126\) rectangles.

Hint

It is tempting to say that there are 11 points and we can choose any three, so \({ }^{11} C_3=165\) possible triangles. Unfortunately, if we choose three points that are on the same row, they will not form a triangle. There are \({ }^5 C_3=10\) ways to select three points on the top row, and \({ }^6 C_3=20\) ways to select three points on the bottom row. When we subtract these from the total, we find that there are 135 triangles that can be formed.

Hint

Every triangle has either point \(A\) or point \(B\) as one vertex. Using \(\mathrm{B}\) as one vertex, we need two more vertices to make a triangle. There are 5 points on segment \(A K\) that can be used as vertices of a triangle with vertex B. We can choose 2 of these in \({ }^5 C_2=10\) ways. The same can be done on AJ, AI, \(\mathrm{AH}\), and \(\mathrm{AG}\) for a total of 50 triangles. Using \(\mathrm{A}\) we have a symmetrical situation in which 50 triangles have one vertex at \(\mathrm{A}\) for a total of \(\mathbf{1 0 0}\) triangles.

Hint

There are \({ }^6 C 3=20\) ways to walk 3 blocks north and three blocks west (consider the number of arrangements of NNNWWW). On the way back, he cannot take the same path, so there are 19 paths to choose for the walk back. This gives us \(20 \cdot 19=\mathbf{3 8 0}\) different ways to walk to the deli and back.

Hint

There are \({ }^{12} C 3=220\) ways to choose three of the twelve points. There are 6 rows of 4 points. If three points are chosen which are on the same line, they cannot be connected to form a triangle. For each row of 4 dots there are \({ }^4 C 3=4\) sets of points which will not form a triangle. This makes a total of \(4 \cdot 6=\) 24 sets which will not form a triangle. \(220-24=\) 196 triangles.

Hint

Every possible pentagon uses one of the three parallel lines on each side. This gives us three choices for each of 5 sides. \(3^5=\mathbf{2 4 3}\) pentagons.

Hint

It is possible to figure out how many ways Paul can flip two, three, four, five, six, seven, or eight heads:
\(
\left(\begin{array}{l}
8 \\
2
\end{array}\right)+\left(\begin{array}{l}
8 \\
3
\end{array}\right)+\left(\begin{array}{l}
8 \\
4
\end{array}\right)+\left(\begin{array}{l}
8 \\
5
\end{array}\right)+\left(\begin{array}{l}
8 \\
6
\end{array}\right)+\left(\begin{array}{l}
8 \\
7
\end{array}\right)+\left(\begin{array}{l}
8 \\
8
\end{array}\right)=247
\)

Alternate way(complementary counting):

It is much easier to figure out the total number of outcomes \(\left(2^8\right)\) and subtract the number of ways he can flip zero or one heads: \({ }^8 C_0=1\) and \({ }^8 C_1=8\).
\(
2^8-1-8=247
\)

Hint

Again, it requires a bit of computation to figure out how many teams include just Corey, just Tony, or both Corey and Tony. Instead, begin by considering how many starting lineups are possible without restrictions (choosing 5 from a team of \(8: { }^8 C_5=56\) ways). Subtract lineups that do not include Corey or Tony (choosing a team of 5 from the remaining 6 players: \({ }^6 C_5=6\) ways). Better still, we can choose one of the 6 remaining players to join Corey and Tony on the bench in \({ }^6 C_1=6\) ways.
\(
\left(\begin{array}{l}
8 \\
5
\end{array}\right)-\left(\begin{array}{l}
6 \\
1
\end{array}\right)=56-6=50 \text { starting lineups. }
\)

Hint

There are \(99-10+1=90\) two-digit integers. 11,22, \(33,44, \ldots 88\), and 99 ( 9 integers) use the same digit twice, so there are \(90-9=81\) two-digit integers that use two different digits.

Hint

There are \({ }^7 C_4=35\) ways to choose four spaces, and only one of these arrangements does not have any spaces next to each other (leaving 34 combinations). Each of these arrangements has 4 ! ways to assign coworkers to the spaces, so we have \(34(4 !)=\mathbf{8 1 6}\) ways to assign the four spaces.

Hint

It is easier to count the ways in which six friends can line up with Alice and David standing together, and subtract them from the total number of ways the students can stand in line without restrictions. There are \(6 !=720\) ways that 6 students can line up. If we pair Alice and David together (Alice immediately to the left of David), there are \(5 !=120\) ways to arrange them. David and Alice can switch places (Alice immediately to the right of David), doubling this number to \(240.720-240=\mathbf{4 8 0}\) ways.

Hint

It is much easier to count the integers less than 50 which are divisible by 3 or 5. There are 16 numbers less than 50 divisible by \(3(3 \cdot 1\) through \(3 \cdot 16)\) and 9 which are divisible by \(5(5 \cdot 1\) through \(5 \cdot 9)\), however, we have counted multiples of 15 twice: 15,30, and 45, so we subtract these from the total. This makes \(16+9-3=22\) numbers less than 50 that are divisible by 3 or 5. There are 49 integers less than 50, so there are \(49-22=\mathbf{2 7}\) positive integers less than 50 which are not divisible by 3 or 5.

Hint

There are \({ }^{10} C 5=252\) ways to flip the same number of heads as tails (choose any 5 of the 10 to be heads), and \(2^{10}=1,024\) ways to flip a coin 10 times. Half of the rest of the times there will be more heads than tails. \((1,024-252) / 2=386\) ways.

Hint

There are \(8^2=64\) possible rolls. It is easier to count rolls which produce a prime number than a composite. A prime product will occur only when one of the dice shows a 1 and the other shows a prime: \((1,2)\) \((1,3)(1,5)\) and \((1,7)\). These four can all be reversed for a total of 8 rolls which have a product that is prime. Finally (and easy to forget), rolling two ones produces a product that is neither prime nor composite, so we have \(64-8-1=55\) rolls.

Hint

When we draw five lines with each intersecting the other four, we get \((5 \cdot 4) / 2=10\) points of intersection (this is a handshake problem). The simplest way to do this is to draw a star. Connecting three of these points to create a triangle requires that we select 3 of the 10 points of intersection \({ }^{10} C_3=120\) ways, however, there are 5 sets of 4 collinear points (selecting three of these will not make a triangle). We must subtract \(5({ }^4 C_3)=20\) sets of points which will not form a triangle, leaving us with \(\mathbf{1 0 0}\) triangles.

Hint

To count the total number of ways to get from \(A\) to \(B\), we must go up 4 times and right 6 times. This gives us \({ }^{10} C_4=210\) ways to get from \(A\) to \(B\). There are \({ }^4 C_2=6\) ways to get from \(\mathrm{A}\) to \(\mathrm{X}\) and \({ }^6 C_2=15\) ways to get from \(X\) to \(B\). This makes \(6 \cdot 15=90\) ways to get from A to B through X. Subtract this from 210 to get \(\mathbf{1 2 0}\) ways which do not pass through \(\mathrm{X}\).

Hint

The gold can be won by = 8 different dogs, after which 7 dogs can win silver, and the remaining 6 can win bronze: 8 x 7 x 6=336 ways.

Hint

PRACTICE has 8 letters with 2 C’s. No of arrangements = 8!/2=20160.

Hint

26 students each complete a project with each of the 25 other students, however, Dhruv working with Calvin is the same as Calvin working with Dhruv. We must divide by 2 to avoid overcounting: 26(25)/2= 325 projects.

Hint

There are 7!=5040 ways to arrange 1st through 7th place. In exactly half of these arrangments, Raj finishes ahead of Thomas. 5040/2=2520 ways.

Hint

There are 5 choices for the 100th digit (1, 3, 5, 7, or 9) then 4 for the 10th digit, and 3 for the one’s digit for a total of 5 x 4 x 3=60 whole numbers.

Hint

Choosing 3 from a group of 15 , there are 15 choices for the first selection, 14 next, and then 13 for the final selection on the team. Picking Al, Bill, and Carly is the same as choosing these three players in any of \(3 !=6\) arrangements.
\({ }^{15} C_3=\frac{15 \cdot 14 \cdot 13}{3 \cdot 2 \cdot 1}=\mathbf{4 5 5}\) possible teams.

Hint

Whether we start with 27 or 28 , there are 7 digits left. There are \(8 \cdot 7 \cdot 6 \cdot 5 \cdot 4 \cdot 3 \cdot 2=40,320\) ways to fill these digits without replacement. 2 choices for the first pair of digits followed by 40,320 possibilities for the remaining digits gives a total of \(\mathbf{8 0 , 6 4 0} \mathrm{NC}\) zip codes that do not repeat digits.

Hint

Three of the roses have already been selected because we know that there must be at least one of each color. This leaves 9 roses to choose from three colors. Place 2 dividers among 9 roses to divide them into red, white, and pink groups. There are \({ }^{11} C_2=55\) ways to place 2 dividers among 9 roses. (There are 11 objects to arrange, 2 are dividers).

Hint

Consider 12 bills and 4 dividers (which separate the bills into five numbered briefcases). There are \({ }^{16} C_4\) \(=\mathbf{1 , 8 2 0}\) ways to arrange 4 dividers among 12 bills. (There are 16 objects to arrange, so we can place the 4 dividers in \({ }^{16} C 4\) ways).

Hint

We place two dividers among 10 points to divide them into groups representing speed, handling, and acceleration. There are \({ }^{12} C_2=\mathbf{6 6}\) ways to arrange 2 dividers among 10 points ( 12 items, 2 are dividers).

Hint

We need \(\left(\begin{array}{l}7 \\ 5\end{array}\right)+\left(\begin{array}{l}7 \\ 6\end{array}\right)+\left(\begin{array}{l}7 \\ 7\end{array}\right)\)

\(21+7+1=29\).

Hint

The 6 slices can be divided among the 4 students who are not vegetarians. Use 3 dividers and 6 slices which gives us \({ }^9 C_3=84\) ways to divide the pepperoni slices. The 3 slices of cheese pizza can be divided among all 5 students. Using 4 dividers and 3 slices we get \({ }^7 C_3=35\) ways to distribute the cheese slices. This gives us a total of \(35 \cdot 84=\mathbf{2 , 9 4 0}\) ways to distribute all 9 slices.

Hint

We are filling the blanks in A _ _ – _ _ _ where the first blank must be filled with one of 24 remaining letters (not A or \(\mathrm{O}\) ), the second blank will be filled with one of 23 letters which remain. The first number can be chosen from among 9 (no 0 ), the second will be one of the remaining 8 , and one of the 7 digits that have not been used will fill the fifth blank. \(24 \cdot 23 \cdot 9 \cdot 8 \cdot 7=\mathbf{2 7 8 , 2 0 8}\) standard license plates.

Hint

No two of the numbers rolled can be the same. There are \({ }^6 C_3=20\) ways to select three numbers (from the six possible) and there is one way to order each of these from least to greatest for white/green/red. This leaves us with just 20 ways.

Hint

\(
{ }^{10} C_2 \cdot { }^{10} C_3=45 \cdot 120=\mathbf{5}, \mathbf{4 0 0} \text { choices. }
\)

Hint

The three couples can be arranged \(3 !=6\) ways on the sofa from left to right, after which each couple can be arranged 2 different ways for a total of \(2^3=8\) ways. After the couples are seated, there are two places for grandma to be seated (between the first two couples or between the \(2^{\text {nd }}\) and \(3^{\text {rd }}\) couple). \(6 \cdot 8 \cdot 2=96\) seating arrangements.

Hint

MISSPELL has 8 letters with two S’s and two L’s. \(\frac{8 !}{2 ! \cdot 2 !}=10,080\) arrangements, but one of these is the correct spelling, so there are \(\mathbf{1 0 , 0 7 9}\) misspellings.