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How many five-card hands are made entirely of “red” cards?

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In mathematics, permutation connects to the process of collecting all the partners of a party into some sequence or format. In different phrases, if the party is already executed, then the redirecting of its members is called the process of permuting. Permutations take place, in more or less significant ways, in nearly every community of mathematics. They often occur when distinct management on specific limited areas is observed.

Permutation

It is the distinct interpretations of a provided number of components carried one by one, or some, or all at a time. For example, if we have two components A and B, then there are two likely performances, AB and BA.

A numeral of permutations when ‘r’ components are positioned out of a total of ‘n’ components is 

nPr = n! / (n – r)!.

For example, let n = 3 (A, B, and C) and r = 2 (All permutations of size 2). The answer is 3!/(3 – 2)! = 6. The six permutations are AB, AC, BA, BC, CA, and CB.

Explanation of Permutation formula

A permutation is a type of performance that indicates how to permute. If there are three different numerals 1, 2, and 3, and if someone is curious to permute the numerals taking 2 at a moment, it shows (1, 2), (1, 3), (2, 1), (2, 3), (3, 1), and (3, 2). That is it can be accomplished in 6 methods.

Here, (1, 2) and (2, 1) are distinct. Again, if these 3 numerals shall be put handling all at a time, then the interpretations will be (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2) and (3, 2, 1) i.e. in 6 ways.

In general, n distinct things can be set taking r (r < n) at a time in n(n – 1)(n – 2)…(n – r + 1) ways. In fact, the first thing can be any of the n things. Now, after choosing the first thing, the second thing will be any of the remaining n – 1 thing. Likewise, the third thing can be any of the remaining n – 2 things. Alike, the rth thing can be any of the remaining n – (r – 1) things.

Hence, the entire number of permutations of n distinct things carrying r at a time is n(n – 1)(n – 2)…[n – (r – 1)] which is written as n Pr. Or, in other words,

nPr = n!/(n – r)!

Combination

It is the distinct sections of a shared number of components carried one by one, or some, or all at a time. For example, if there are two components A and B, then there is only one way to select two things, select both of them.

Number of combinations when ‘r’ components are chosen out of a total of ‘n’ components is, 

nCr = n!/[(r!) × (n – r)!]

For example, let n = 3 (A, B, and C) and r = 2 (All combinations of size 2). The answer is 3!/((3 – 2)! × 2!) = 3. The six combinations are AB, AC, and BC.

nCr = nC(n-r)

Note: In the same example, we have distinct points for permutation and combination. For, AB and BA are two distinct items but for selecting, AB and BA are the same.

Explanation of Combination Formula

Combination, on the further hand, is a type of pack. Again, out of those three numbers 1, 2, and 3 if sets are created with two numbers, then the combinations are (1, 2), (1, 3), and (2, 3).

Here, (1, 2) and (2, 1) are identical, unlike permutations where they are distinct. This is written as 3C2. In general, the number of combinations of n distinct things taken r at a time is,

nCr = n! /[r! × (n – r)!] = nPr/r!

How many five-card hands are made entirely of “red” cards? 

Solution:

There are a total of 26 red cards i.e., 13 hearts and 13 diamonds.

From 26 red cards, choose 5.

The answer is the binomial coefficient

(26C5) and you can read this as 26 choose 5.

So there are

(26C5) = 26! ⁄ 5!(26−5)!  

= 26! ⁄ 5!21!  

= 26×25×24×23×22×21! ⁄ 5×4×3×2×1×21!  

= 26×25×24×23×22 ⁄ 5×4×3×2  

= 26⁄2×25⁄5×24⁄12×23×22  

= 13×5×2×23×22  

= 13×10×23×22  

= 130×506 = 65,780

Possible 5-card hands consisting of only red cards.

Similar Problems

Question 1: From a standard 52 card deck, how many 4 card hands consist entirely of black cards?

Solution:

There is a total of 26 black cards i.e., 13 clubs and 13 spades.

From 26 black cards, choose 4.

The answer is the binomial coefficient

(26C4) and you can read this as 26 choose 4.

So there are

(26C4) = 26! ⁄ 4!(26−4)!  

= 26! ⁄ 4!22!  

= 26×25×24×23×22! ⁄ 4×3×2×1×22!  

= 26×25×24×23 ⁄ 4×3×2

= 358,800/24

= 14,950

Possible 4-card hands consisting of only black cards.

Question 2: How many different arrangements of the word ‘MATHEMATICS’ are possible?

Solution:

11 letters but no repetition

(4 M’s, 2 A’s, 2 T’s, and others H, E, I, C, S are 1 each.)

No. of arrangements =11!/2! 2! 2! single alphabets ignored

= 4989600

Question 3: There are 4 boys who enter a boat with 6 seats, 3 on each side. In how many ways can:

a) they sit anywhere?

Solution: 

6P4

=6!/(6-4)! 

=6!/2!

=360

b) two boys Q and R sit on the dockside and another boy L sit on the starboard side?

Solution:

A and B= 3P2

W = 3P1

Others = 3P1

Total = 3P2 × 3P1 × 3P

= 6×3×3

= 54

Question 4: At a lunch party, 4 men and 4 women sit around the table. In how many ways can they sit if:

a) Jaya and Diya must sit together

Solution: (JD) and other 6 = 2! × 6!

                                           = 1440

b) Jaya, Diya, and Maya must sit together

Solution: (JDM) and other 5 = 3! × 5!

                                              = 720

Question 5: In how many ways can 8 differently colored dots be knitted on a row?

Solution:

As row can be turned over, clockwise and anti-clockwise arrangements are the same

= (8-1)! / 2 

= 7! / 2

=2520


Last Updated : 30 Jan, 2022
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