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## Comment on

Handling Restrictions## 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

## We can arrange 10 students in

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!

## We can arrange 10 students in

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

## How will you solve this

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?

## Great question.

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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