Respuesta :
Answer:
Total number of cards in hand contain exactly two 3s and two 7s is 56.
Step-by-step explanation:
Given : This exercise refers to a standard deck of playing cards. Assume that 5 cards are randomly chosen from the deck.
To find : How many hands contain exactly two 3s and two 7s?
Solution :
In a standard deck of card has 52 cards distributed in 4 sets.
Each set has {2,3,4,5,6,7,8,9,10,J,Q,K,A} has 13 cards.
Number of 3s in 52 cards are 4.
Number of 7s in 52 cards are 4.
So, Getting exactly two 3s out of 4 is [tex]^4C_2[/tex]
Getting exactly two 7s out of 4 is [tex]^4C_2[/tex]
The fifth card is from remaining cards i.e. 52-8=44
Getting fifth card is [tex]^{44}C_1[/tex]
Total number of cards in hand contain exactly two 3s and two 7s is
[tex]T=^4C_2+^4C_2+^{44}C_1[/tex]
[tex]T=\frac{4!}{2!(4-2)!}+\frac{4!}{2!(4-2)!}+\frac{44!}{1!(44-1)!}[/tex]
[tex]T=\frac{4\times3\times2!}{2!\times 2}+\frac{4\times3\times2!}{2!\times 2}+\frac{44\times43!}{1\times 43!}[/tex]
[tex]T=6+6+44[/tex]
[tex]T=56[/tex]
Therefore, Total number of cards in hand contain exactly two 3s and two 7s is 56.
The question is an illustration of combination (also known as selections)
The number of hands that contain exactly two 3s and two 7s is 1584
From the complete question:
- Two 3s are to be selected from 4
- Two 7s are to be selected from 4
Select two 3s from 4
This means that:
- [tex]\mathbf{n =4}[/tex]
- [tex]\mathbf{r =2}[/tex]
So, the number of selection is:
[tex]\mathbf{^4C_2=\frac{n!}{(n - r)!r!} }[/tex]
[tex]\mathbf{^4C_2=\frac{4!}{(4 - 2)!2!} }[/tex]
[tex]\mathbf{^4C_2=\frac{4!}{2!2!} }[/tex]
[tex]\mathbf{^4C_2=\frac{4 \times 3 \times 2!}{2! \times 2\times 1} }[/tex]
[tex]\mathbf{^4C_2=6 }[/tex]
Select two 7s from 4
This means that:
- [tex]\mathbf{n =4}[/tex].
- [tex]\mathbf{r =2}[/tex]
So, the number of selection is:
[tex]\mathbf{^4C_2=\frac{n!}{(n - r)!r!} }[/tex]
[tex]\mathbf{^4C_2=6 }[/tex]
The number of cards in a standard deck is 52.
So, the remaining card is:
[tex]\mathbf{n =52 - 4 - 4}[/tex]
[tex]\mathbf{n =44}[/tex]
Also, the number of remaining card to be selected is:
[tex]\mathbf{r = 5 -2-2}[/tex]
[tex]\mathbf{r =1}[/tex]
So, we have:
[tex]\mathbf{n =44}[/tex]
[tex]\mathbf{r =1}[/tex]
So, the number of last selection is:
[tex]\mathbf{^44C_1=\frac{44!}{(44 - 1)!1!} }[/tex]
[tex]\mathbf{^44C_1=\frac{44!}{43! \times 1!} }[/tex]
[tex]\mathbf{^44C_1=44}[/tex]
So, the total selection is:
[tex]\mathbf{Total = 6 \times 6 \times 44}[/tex]
[tex]\mathbf{Total = 1584}[/tex]
Hence, the number of hands that contain exactly two 3s and two 7s is 1584
Read more about combinations at:
https://brainly.com/question/15301090
