18 mice were placed in two experimental groups and one control group, with all groups equally large. In how many ways can the mice be placed into three groups?
The total number of mice is 18 and they must be divided into three equal groups of 6 each.
The number of ways to divide \(18\) distinct mice into groups of size \(6, 6, 6\) is given by:
\[ \dfrac{18!}{(6!)^3} \]
Thus, the required number of ways is \( \dfrac{18!}{(6!)^3} \).
A bag contains six white marbles and five red marbles. Find the number of ways in which four marbles can be drawn from the bag if:
(a) they can be of any colour
(b) two must be white and two red
(c) they must all be of the same colour
(a) Total marbles = 11. Number of ways to draw 4 of any colour:
\[ \binom{11}{4} \]
(b) Two white and two red:
\[ \binom{6}{2} \times \binom{5}{2} \]
(c) All marbles same colour:
All white: \( \binom{6}{4} \)
All red: \( \binom{5}{4} \)
Total: \( \binom{6}{4} + \binom{5}{4} \)
In how many ways can a football team of 11 players be selected from 16 players? How many of them will:
(i) include 2 particular players?
(ii) exclude 2 particular players?
Total ways to select 11 players from 16:
\[ \binom{16}{11} \]
(i) If 2 particular players are included, then we choose the remaining 9 from the remaining 14:
\[ \binom{14}{9} \]
(ii) If 2 particular players are excluded, then we select all 11 from the remaining 14:
\[ \binom{14}{11} \]
A sports team of 11 students is to be constituted, choosing at least 5 from Class XI and at least 5 from Class XII. If there are 20 students in each of these classes, in how many ways can the team be constituted?
The team must contain:
Case 1: 5 from Class XI and 6 from Class XII
Case 2: 6 from Class XI and 5 from Class XII
Total ways:
\[ \binom{20}{5} \times \binom{20}{6} + \binom{20}{6} \times \binom{20}{5} \]
Since the two terms are identical, the total is:
\[ 2 \left( \binom{20}{5} \times \binom{20}{6} \right) \]
A group consists of 4 girls and 7 boys. In how many ways can a team of 5 members be selected if the team has:
(i) no girls
(ii) at least one boy and one girl
(iii) at least three girls
(i) No girls means all 5 selected from 7 boys:
\[ \binom{7}{5} = 21 \]
(ii) At least one boy and one girl:
Total ways to choose 5 from 11:
\[ \binom{11}{5} \]
Subtract teams with all boys and all girls:
All boys: \( \binom{7}{5} \)
All girls: \( \binom{4}{5} = 0 \)
Thus, ways =
\[ \binom{11}{5} - \binom{7}{5} = 462 - 21 = 441 \]
(iii) At least 3 girls:
Possible combinations:
3 girls + 2 boys: \( \binom{4}{3} \binom{7}{2} \)
4 girls + 1 boy: \( \binom{4}{4} \binom{7}{1} \)
Total:
\[ \binom{4}{3} \binom{7}{2} + \binom{4}{4} \binom{7}{1} = 4 \times 21 + 1 \times 7 = 84 + 7 = 91 \]