In how many ways a hockey team of eleven can be elected from 16 players ?
This problem involves selecting a team from a larger group of players, where the order of selection does not matter. This is a classic combination problem. We need to find the number of ways to choose 11 players from 16 players.
Identify the type of problem: Since the order in which players are chosen for the team does not matter (a team of players A, B, C is the same as a team of B, A, C), this is a combination problem.
Recall the combination formula: The number of ways to choose \(k\) items from a set of \(n\) items (where order does not matter) is given by the combination formula:
\[ \binom{n}{k} = C(n, k) = \frac{n!}{k!(n-k)!} \]where \(n!\) (n factorial) is the product of all positive integers up to \(n\).
Identify \(n\) and \(k\):
Apply the formula: Substitute the values of \(n\) and \(k\) into the combination formula:
\[ C(16, 11) = \frac{16!}{11!(16-11)!} = \frac{16!}{11!5!} \]Calculate the factorials and simplify:
\[ C(16, 11) = \frac{16 \times 15 \times 14 \times 13 \times 12 \times 11!}{11! \times 5 \times 4 \times 3 \times 2 \times 1} \]Cancel out \(11!\) from the numerator and denominator:
\[ C(16, 11) = \frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1} \]Simplify the denominator: \(5 \times 4 \times 3 \times 2 \times 1 = 120\)
\[ C(16, 11) = \frac{16 \times 15 \times 14 \times 13 \times 12}{120} \]Further simplification:
Let's simplify step by step:
\[ C(16, 11) = \frac{16 \times 15 \times 14 \times 13 \times 12}{5 \times 4 \times 3 \times 2 \times 1} \]So, the expression becomes:
\[ C(16, 11) = 4 \times 1 \times 14 \times 13 \times 6 \] \[ C(16, 11) = 4 \times 14 \times 13 \times 6 \] \[ C(16, 11) = 56 \times 78 \] \[ C(16, 11) = 4368 \]Correct Option: C) 4368
Three dice are rolled. The number of possible outcomes in which at least one die shows 5 is
To find the number of outcomes where at least one die shows 5, it's easier to calculate the total number of possible outcomes and subtract the number of outcomes where no die shows 5. This is based on the principle of complementary counting.
When three dice are rolled, each die has 6 possible outcomes (1, 2, 3, 4, 5, 6).
\[ \text{Total outcomes} = 6 \times 6 \times 6 = 6^3 = 216 \]If no die shows 5, then each die can show any number from {1, 2, 3, 4, 6}. This means each die has 5 possible outcomes.
\[ \text{Outcomes with no 5} = 5 \times 5 \times 5 = 5^3 = 125 \]Using the principle of complementary counting:
\[ \text{At least one 5} = \text{Total outcomes} - \text{Outcomes with no 5} \]\[ \text{At least one 5} = 216 - 125 = 91 \]Thus, there are 91 possible outcomes in which at least one die shows 5.
B) 91 is the correct answer, derived by subtracting the outcomes with no 5 from the total outcomes.
The number of three digit numbers having only two consecutive digits identical is
We need to find the number of three-digit numbers where exactly two consecutive digits are identical. A three-digit number can be represented as \(abc\), where \(a\) is the hundreds digit, \(b\) is the tens digit, and \(c\) is the units digit. The condition "only two consecutive digits identical" means either \(a=b\) and \(b \neq c\), or \(b=c\) and \(a \neq b\).
The two cases are mutually exclusive (a number cannot be of the form \(aac\) and \(abb\) simultaneously unless all three digits are the same, which is excluded by the condition "only two consecutive digits identical").
Total = (Numbers from Case 1) + (Numbers from Case 2)
\[ \text{Total} = 81 + 81 = 162 \]B) 162
A lady gives a dinner party to 5 guests to be selected from nine friends. The number of ways of forming the party of 5, given that two of the friends will not attend the party together is
This problem involves combinations with a restriction. We need to select 5 guests from 9 friends, with the condition that two specific friends will not attend the party together. We can solve this by considering cases: either neither of the two specific friends is invited, or one of them is invited, but not both.
B) 91 is the correct answer, as calculated by summing the ways from the two mutually exclusive cases that satisfy the given condition.
In how many ways can four children be made to stand in a line such that two of them, A and B are always together ?
This problem involves permutations with a constraint. We need to find the number of ways to arrange four children in a line such that two specific children, A and B, are always together. The key idea is to treat the two children who must be together as a single unit.
B) 12 is the correct answer. There are 12 ways to arrange the four children such that A and B are always together.
There are 10 lamps in a hall. Each of them can be switched on independently. Find the number of ways in which the hall can be illuminated.
This problem can be solved using the fundamental principle of counting. Each lamp has two independent states: either it is ON or it is OFF. We need to find the total number of combinations of these states such that the hall is illuminated, meaning at least one lamp must be ON.
Determine the number of states for each lamp: Each lamp can be either ON or OFF. So, for each lamp, there are 2 possible states.
Calculate the total number of ways to switch the lamps: Since there are 10 lamps and each can be independently in 2 states, the total number of ways to switch all 10 lamps is \(2 \times 2 \times \dots \times 2\) (10 times), which is \(2^{10}\).
\[ 2^{10} = 1024 \]This total includes the case where all lamps are OFF.
Identify the condition for illumination: The hall is illuminated if at least one lamp is ON. This means we must exclude the case where all lamps are OFF.
Subtract the undesirable case: There is only one way for all lamps to be OFF (Lamp 1 OFF, Lamp 2 OFF, ..., Lamp 10 OFF).
Calculate the number of ways to illuminate the hall: Subtract the case where all lamps are OFF from the total number of ways to switch the lamps.
\[ \text{Number of ways to illuminate} = \text{Total ways} - \text{Ways with all lamps OFF} \] \[ \text{Number of ways to illuminate} = 2^{10} - 1 \] \[ \text{Number of ways to illuminate} = 1024 - 1 \] \[ \text{Number of ways to illuminate} = 1023 \]Correct Option: A) 1023
