Question

in how many ways can a group consisting of 5 girls and 4 boys be formed from a class of 8 girls and 5 boys​

Answer 1

Answer:

280 ways with a Probability of 39.12%

280 ways with a Probability of 39.12%5 girls to be picked up from total number of 8 girls

280 ways with a Probability of 39.12%5 girls to be picked up from total number of 8 girlsNumber of ways pick up girls is 8 C 5 =[ (8 !)/(5!) (3!)] = [ (8 x 7 x 6)/ (3 x 2 x 1)] = 56

280 ways with a Probability of 39.12%5 girls to be picked up from total number of 8 girlsNumber of ways pick up girls is 8 C 5 =[ (8 !)/(5!) (3!)] = [ (8 x 7 x 6)/ (3 x 2 x 1)] = 564 Boys to be picked up from total number of 5 Boys

280 ways with a Probability of 39.12%5 girls to be picked up from total number of 8 girlsNumber of ways pick up girls is 8 C 5 =[ (8 !)/(5!) (3!)] = [ (8 x 7 x 6)/ (3 x 2 x 1)] = 564 Boys to be picked up from total number of 5 BoysNumber of ways pick up boys is 5 C 4 =[ (5 !)/(4!) (1!)] = [ ( 5 )/ (1)] = 5

280 ways with a Probability of 39.12%5 girls to be picked up from total number of 8 girlsNumber of ways pick up girls is 8 C 5 =[ (8 !)/(5!) (3!)] = [ (8 x 7 x 6)/ (3 x 2 x 1)] = 564 Boys to be picked up from total number of 5 BoysNumber of ways pick up boys is 5 C 4 =[ (5 !)/(4!) (1!)] = [ ( 5 )/ (1)] = 5Hence, the total number of ways to pick up 5 Boys and 8 Girls = (56 x 5) = 280

280 ways with a Probability of 39.12%5 girls to be picked up from total number of 8 girlsNumber of ways pick up girls is 8 C 5 =[ (8 !)/(5!) (3!)] = [ (8 x 7 x 6)/ (3 x 2 x 1)] = 564 Boys to be picked up from total number of 5 BoysNumber of ways pick up boys is 5 C 4 =[ (5 !)/(4!) (1!)] = [ ( 5 )/ (1)] = 5Hence, the total number of ways to pick up 5 Boys and 8 Girls = (56 x 5) = 280If we calculate the Probability, the total number of 9 students picking up from 13 students is [( 13!) / (9 ! ) (4 !)] = [ (13 x 12 x 11 x 10 )/( 4 x 3 x 2 x 1)] = 715

280 ways with a Probability of 39.12%5 girls to be picked up from total number of 8 girlsNumber of ways pick up girls is 8 C 5 =[ (8 !)/(5!) (3!)] = [ (8 x 7 x 6)/ (3 x 2 x 1)] = 564 Boys to be picked up from total number of 5 BoysNumber of ways pick up boys is 5 C 4 =[ (5 !)/(4!) (1!)] = [ ( 5 )/ (1)] = 5Hence, the total number of ways to pick up 5 Boys and 8 Girls = (56 x 5) = 280If we calculate the Probability, the total number of 9 students picking up from 13 students is [( 13!) / (9 ! ) (4 !)] = [ (13 x 12 x 11 x 10 )/( 4 x 3 x 2 x 1)] = 715Hence the Probability of forming a group of 5 girls and 4 boys from a group of 8 girls and 5 boys is (280/715) = 39.12 %

Hope it helps uh

Answer 2

Answer:

The number of ways in which 5 girls and 4 boys can be formed from a class of 8 girls and 5 boys = 280 ways

Step-by-step explanation:

Required to find,

The number of ways in which 5 girls and 4 boys can be formed from a class of 8 girls and 5 boys.

Recall the concepts

The number of combinations of 'n' distinct objects taken 'r' at a time is given by $nC_r$  

$nC_r = \frac{n!}{r!(n-r)!} , 0\leq r\leq n$ -------------(1)

$nC_r = nC_{n-r}$ --------------------------(2)

Solution:

Out of 8 girls, 5 girls can be chosen in $8C_5$ ways

Out of 5 boys, 4 boys can be chosen in $5C_4$ ways

Hence, out of 8 girls and 5 boys, 5 girls and 4 boys can be chosen in

$8C_5$ ×$5C_4$ ways

From formula (2), we get $8C_5 =8C_3$

From formula (1), we get   $8C_5 = 8C_3 = \frac{8!}{5! 3!} = \frac{8X7X6}{1X2X3} = 56$

From formula (2), we get $5C_4 = 5C_1$

From formula (1), we get  $5C_4 = 5C_1 = 5$

∴  $8C_5 X5C_4$ = 56 × 5 = 280

The number of ways in which 5 girls and 4 boys can be formed from a class of 8 girls and 5 boys = 280 ways

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