Question

A standard deck of cards consists of four suits (clubs, diamonds, hearts, and spades), with each suit containing 13 cards (ace, two through ten, jack, queen, and king) for a total of 52 cards in all. How many 7-card hands will consist of exactly 2 hearts and 3 clubs?

Answer 1

Answer:

1,977,300

Step-by-step explanation:

Given,

Total cards = 52,

In which,

Clubs = 13, diamond = 13, heart = 13, spades = 13,

Ways of choosing exactly 2 hearts = $^{13}C_2$

Ways of choosing exactly 3 clubs = $^{13}C_3$

So, the total possible ways of selecting 7 cards that contains exactly 2 hearts and 3 clubs

= Choosing exactly 2 hearts × Choosing exactly 3 clubs × choosing any 2 cards except hearts and clubs

$=^{13}C_2\times ^{13}C_3\times ^{26}C_2$

$=\frac{13!}{2!11!}\times \frac{13!}{2!11!}\times \frac{26!}{2!24!}$

$=78\times 78\times 325$

= 1,977,300

Answer 2

Using the combination formula, it is found that 7,250,100 7-card hands will consist of exactly 2 hearts and 3 clubs.

--------------------

Combination formula:

The number of different combinations of x objects from a set of n elements is given by:

$C_{n,x} = \frac{n!}{x!(n-x)!}$

In this question:

• 2 hearts, from a set of 13.
• 3 clubs, from a set of 13.
• 2 diamonds or spades, from a set of 26.

Then

$T = C_{13,2}C_{13,3}C_{26,2} = \frac{13!}{2!11!} \times \frac{13!}{3!10!} \times \frac{26!}{2!24!} = 78 \times 286 \times 325 = 7,250,100$

7,250,100 7-card hands will consist of exactly 2 hearts and 3 clubs.

A similar problem is given at https://brainly.com/question/24437717