Lesson: Handling Restrictions

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Dear Sir

IF the total number of ways in which 10 students can be arranged in a row such that A is always ahead of B?

a. 2x10!
b. 10! /2
c. 10! x 8!
d. none

Regards
greenlight-admin's picture

We can arrange 10 students in 10! ways.

In HALF of those 10! arrangements, A will be ahead of B, and in the other HALF of those 10! arrangements, B will be ahead of A

So, the number of arrangements where A is ahead of B is 10!/2 (and the number of arrangements where B is ahead of A is 10!/2)

I didnt understand how in 10!/2 arrangement A will be ahead of B, please explain?
greenlight-admin's picture

We can arrange 10 students in 10! ways.

In each arrangement, EITHER A is ahead of B OR B is ahead of A.

Since the there's no reason why one student (A or B) should be ahead of the other student, we can say that, in HALF of the arrangements A will be ahead of B, and in the other HALF of those 10! arrangements, B will be ahead of A.

As Sir explained that A has to be ahead of B so it does not mean that they have to be together too. A can be two ahead of B too . Similarly three steps. So if there are 10! factorial ways in total , then there will be definitely half of the ways when A is ahead and Half of ways when B is ahead.

How will you solve this problem in the context of unique arrangements?

A card game using 36 unique cards, four suits(diamonds, hearts, clubs and spades) with cards numbered from 1-9 in each suit. A hand is a collection of 9 cards, which can be sorted however the player chooses. What is the probabilities of getting all four of the 1s?
greenlight-admin's picture

Great question.

P(getting all four 1's) = (total number of ways to select 9 cards that include all four 1's)/(total number of ways to select 9 cards)

DENOMINATOR first: Total number of ways to select 9 cards.
We can select 9 cards from 36 cards in 36C9 ways
36C9 = 94,143,280

NUMERATOR second: since the four 1's must be in the hand, let's place those four 1's in the hand right away, and then select the remaining 5 cards. Once we place the four 1's in the hand, there are 32 cards remaining. We can select 5 cards from 32 cards in 32C5 ways
32C5 = 201,376

So, P(getting all four 1's) = 201,376/94,143,280
≈ 0.00214

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