A

9000139506

Level: 
A
There are eight mandarins of average mass \(90\, \mathrm{g}\) in the box. We got two another mandarins and add them to the box. The new average mass of the mandarins in the box is \(92\, \mathrm{g}\). Find the average mass of the two added mandarins.
\(100\, \mathrm{g}\)
\(92\, \mathrm{g}\)
\(96\, \mathrm{g}\)
\(106\, \mathrm{g}\)

9000139303

Level: 
A
A DJ's playlist contains \(18\) songs. In this list there are \(7\) rap songs, \(5\) oldies and \(6\) rock songs. The opening part should consist of one rap song, two oldies and one rock song. The order of the songs does not matter. Find the number of possible ways how to put the opening together.
\(420\)
\(120\)
\(320\)
\(520\)

9000139703

Level: 
A
The box contains \(5\) red crayons, \(4\) yellow crayons and \(2\) green crayons. The crayons are removed from the box and arranged in a line. How many different color patterns can be obtained by this procedure?
\(\frac{11!} {5!\, 4!\, 2!}=6\:930\)
\(5\cdot 4\cdot 2=40\)
\(5!\, 4!\, 2!=5\:760\)
\(\left (5!\, 4!\right )^{2}=8\:294\:400\)

9000139305

Level: 
A
There are five rooms with three beds and one room with five beds in a hotel. A group of \(20\) people booked rooms at this hotel. Determine the number of possible choices for the people to the five-bed room.
\(\frac{20!} {5!\; 15!}=15\:504\)
\(20\cdot 3\cdot 5=300\)
\(\frac{20!} {3!\; 5!}=3\:379\:030\:566\:912\:000\)
\(20^{5}=3\:200\:000\)

9000139705

Level: 
A
From the group of \(10\) boys and \(5\) girls we have to select a small group of \(3\) boys and \(2\) girls. How many possibilities exist for this choice?
\(\frac{10!} {7!\, 3!}\cdot \frac{5!} {3!\, 2!}=1\:200\)
\(5^{10}=9\:765\:625\)
\(10\cdot 5!\, 3!=7\:200\)
\(5\cdot \frac{10!} {3!} =3\:024\:000\)

9000139706

Level: 
A
The international alphabet contains \(26\) letters. The letters of this alphabet and the digits from \(0\) to \(9\) are used to form a code of the length \(4\) (a code contains \(4\) characters). The characters may repeat through the code and the code is not case sensitive (uppercase letters are equivalent to lowercase letters). How many codes can be obtained?
\(36^{4}=1\:679\:616\)
\(10\cdot 26^{4}=4\:569\:760\)
\(\frac{36!} {32!\, 4!}=58\:905\)
\(\frac{26!} {22!\, 4!}=14\:950\)

9000139302

Level: 
A
The phone number contains nine digits. A witness does not remember the full number, but he remembers that the phone number starts by \(728\), ends by \(01\) and there is no repeating digit in the number. How many phone numbers meet these conditions?
\(120\)
\(320\)
\(520\)
\(720\)