Factorial:
The factorial of a number n is the product of all natural numbers less than or equal to n i.e. n! = 1 * 2 * 3 * . . . * (n-1) * n
Also, remember that 0! =1
Selection without arrangement:
Choosing r out of n options can be represented as
$$^n C _r$$ = $$\frac{n!}{r!(n-r)!}$$ .
If we have to choose a team of 11 players out of 15 available players, we can do that in $$ ^{15} C _{11} $$ ways = 15!/(11!4!) = 1365 ways. Hence 11 players can be chosen in 1365 ways.
Consider the earlier example of picking 11 players for the team out of 14. If suppose there are two players Ajay and Vijay among them who are best friends and always want to play together, we have the added condition that if one of Ajay/ Vijay is picked the other one must always be picked.
To find the number of ways in this condition, we consider two cases - one case where neither of the two are picked and one case where both are picked. In the first case, we have only 15-2 =13 options to pick from. Hence, the team can be selected in $$^{13} C _{11}$$ ways = 78 ways. In the second case, since 2 players are already selected we just have to pick the remaining 9 players. Hence, the team can be selected in $$^{13} C _{9}$$ = 715 ways. Hence, the total number of ways of selecting the team with this additional condition is 715+78=793 ways.
Suppose if Ajay and Vijay from the earlier example did not get along and could not picked together. In this case, we can consider three cases - one when Ajay is picked, one when Vijay is picked and one when neither is picked. The sum of these three cases is equal to the number of cases without condition - cases where they are picked together. Number of ways they can be picked together is 715 ways as seen earlier. Hence, number of ways in which at max only one of them is picked = 1365 - 715= 650 ways.
Consider you have n distinct items. They can be arranged among themselves in n! ways.
Consider that you have n distinct items out of which you have to pick r items and arrange them. This can be done in
$$^n P _r$$ ways = $$\frac{n!}{(n-r)!}$$.
There are 5 professors in the Economics department. Suppose we had to schedule three lectures - one in the morning, one in afternoon and one in evening from these 5 professors. We can do this in $$^5 P _3$$ ways = 5!/2! = 60 ways.
If the arrangement should always begin or end with a particular item, its order is fixed. Hence the remaining n-1 items can be arranged in (n-1)! ways.
Without additional conditions n distinct items can be arranged in n! ways. However, if we have the additional constraint that 2 items must always be together. For example, if we are arranging 10 students in a photo such that two students should always be together. In such cases, consider the two to be a single unit. Hence, the other 10-2 students and this unit can be arranged in (10-2+1)!= 9! ways. At the same time, the two can be arranged in 2! ways within the unit. Hence the students can be arranged in 2!*9! ways.
If n items are to arranged in a line of which r items must never be together, we can arrange them in
No. of ways= (n-r)!r! $$^{n-r+1} C _r$$
provided n-r is greater than r
10 students are standing in a line to get their photograph taken. 2 out of these 10 students can never stand together. In how many ways can the students be arranged.
In such cases first lets arrange the remaining 8 students. They can be arranged in 8! ways. Since they are arranged in a straight line this can be represented in x S x S x S x S x S x S x S x S x where S are the eight students and x is the space to the left and right of these students. The remaining 2 students can be arranged in any of these 9 spaces. We can choose two spaces in $$^9 C _2$$ ways and arrange the two among themselves in 2! ways. Hence total number of ways of arranging the students = 8! * $$^9 C _2$$ * 2! = 72 * 8!
• Probability of an event: If in n equally likely outcomes an event is said to have occurred m times, then the probability of the event is P(E) = m/n
• Probability of the event not occurring is P (E') = 1-m/n
• Events are said to be mutually exclusive if they can never occur together. They are said to be collectively exhaustive if together they encompass all possible outcomes.
• The sum of probabilities of mutually exclusive and exhaustive events is 1
• Probability of event A or B occurring P (A or B) = P(A) + P(B) – P(A and B)
How many 4-digit numbers with distinct digits are not divisible by 2?
Explanation:
The first digit cannot be zero. Let the last digit be 1. The other three digits can be selected in 8*8*7 = 448 ways. If the last digit is 3, another 448 ways. Similarly for 5, 7 and 9. The total number of ways of selection is 448*5 = 2240 ways.
The probabilities of solving a question by three students P, Q, R are 1/3, 2/5 and ½ respectively. The probability that the question is solved is?
Explanation:
P (question is solved) = 1 – P (no student solves the question) = 1 – 2/3*3/5*1/2 = 1 – 1/5 = 4/5.
