Question

A farmer buys 3 cows, 2 pigs, and 4 hens from a man who has 6 cows, 5 pigs, and 8 hens. find the number m of choices that the farmer has.

Answer 1

The farmer has 14000 choices

Correct question : A farmer buys 3 cows, 2 pigs, and 4 hens from a man who has 6 cows, 5 pigs, and 8 hens. find the number of choices that the farmer has.

Given :

A farmer buys 3 cows, 2 pigs, and 4 hens from a man who has 6 cows, 5 pigs, and 8 hens.

To find :

The number of choices that the farmer has.

Solution :

Step 1 of 2 :

Calculate number of choices for each of cows, pigs, hens

Here it is given that the farmer buys 3 cows, 2 pigs, and 4 hens from a man who has 6 cows, 5 pigs, and 8 hens.

Farmer can choose 3 cows from 6 cows in ⁶C₃ ways

Farmer can choose 2 pigs from 5 pigs in ⁵C₂ ways

Farmer can choose 4 hens from 8 hens in ⁸C₄ ways

Step 2 of 2 :

Calculate the total number of choices that the farmer has.

The number of choices that the farmer has

$\displaystyle \sf{ = {}^{6}C_3 \times {}^{5}C_2 \times {}^{8}C_4 }$

$\displaystyle \sf{ = \frac{6!}{3! \: (6 - 3)!} \times \frac{5!}{2! \: (5 - 2)!} \times \frac{8!}{4! \: (8 - 4)!} }$

$\displaystyle \sf{ = \frac{6!}{3! \: 3!} \times \frac{5!}{2! \: 3!} \times \frac{8!}{4! \: 4!} }$

$\displaystyle \sf{ = \frac{6 \times 5 \times 4 \times 3!}{3! \: 3!} \times \frac{5 \times 4 \times 3!}{2! \: 3!} \times \frac{8 \times 7 \times 6 \times 5 \times 4 !}{4! \: 4!} }$

$\displaystyle \sf{ = \frac{6 \times 5 \times 4 }{3! } \times \frac{5 \times 4 }{2! } \times \frac{8 \times 7 \times 6 \times 5 }{4! } }$

$\displaystyle \sf{ = \frac{6 \times 5 \times 4 }{3 \times 2 \times 1} \times \frac{5 \times 4 }{2 \times 1 } \times \frac{8 \times 7 \times 6 \times 5 }{4 \times 3 \times 2 \times 2 } }$

$\displaystyle \sf{ = 20 \times 10 \times 70 }$

$\displaystyle \sf{ = 14000 }$

━━━━━━━━━━━━━━━━

Learn more from Brainly :-

$1.$ If 7. 5pr = 7pr – 1, then find r.(question based on permutation and combination chapter)

https://brainly.in/question/26877723

2. If nPr = 3024 and nCr = 126 then find n and r.

https://brainly.in/question/29044585

#SPJ3

Answer 2

Answer:

The number of choices that the farmer has 14000.

Step-by-step explanation:

Explanation:

Given, farmer buys 3 cows , 2 pigs and 4 hens from a man who has 6 cows , 5 pigs and 8 hens.

This means that, farmer buys 3 cows , 2 pigs and 4 hens

and the man has 6 cows , 5 pigs and 8 hens.

Step 1:

Number of choices that the farmer has  for cows = $6C_3$

Number of choices that the farmer has for pigs = $5C_2$

Number of choices that the farmer has for hens = $8 C_4$

Therefore, the number of choices that the farmer has ,

$6C_3 .5C_2. 8C_4$

⇒ $\frac{6!}{2!(6-2)!}$ × $\frac{5!}{2!(5-2)!}$ ×$\frac{8!}{4!(8-4)!}$

⇒$\frac{6!}{3! 3!}$ × $\frac{5!}{2! 3!}$ × $\frac{8!}{4! 4!}$

⇒$\frac{6 . 5 . 4 . 3!}{3!3!}$ ×  $\frac{5.4.3!}{2!3!}$ × $\frac{8.7.6.5.4!}{4!4!}$

⇒$\frac{6.5.4}{3 .2.1}$ × $\frac{5.4}{2.1}$ × $\frac{8.7.6.5}{4.3.2.1}$

⇒ 20 × 10 × 70

⇒14000

Final answer:

Hence, the number of choices that the farmer has 14000.

#SPJ2