Question

A committee of eight congressman will be selected from a group of 11 Democrats and seven Republicans. Find the number of ways of obtaining a committee with exactly 3 Democrats.

Answer 1

Answer:

$3465$

Step-by-step explanation:

GIVEN: A committee of eight congressman will be selected from a group of $11$ Democrats and $7$ Republicans.

TO FIND: The number of ways of obtaining a committee with exactly $3$Democrats.

SOLUTION:

Total number of congressman in committee $=8$

Total number of republicans available $=7$

Total number of democrats available $=11$

As, the committee should have exactly $3$ democrats, there will be $5$ republicans in committee

No. of ways of selecting $3$ democrats out of $11$ democrats $^1^1C_3=\frac{11!}{(11-3)!3!}$

                                                                                               $=\frac{11!}{8!3!}$

                                                                                               $=165$

NO. of ways of selecting $5$ republicans out of $7$ republicans  $^7C_5=\frac{7!}{(7-5)!5!}$

                                                                                                   $=\frac{7!}{5!2!}$

                                                                                                   $=21$

The number of ways of obtaining a committee with exactly $3$ Democrats.

                  $=\text{total ways of selecting 3 democrats}\times\text{total ways of selecting 5 republicans}$

$=165\times21$

$=3465$

Hence there are $3465$ ways to form a committee in which there are exactly $3$ democrats.

Answer 2

Answer: 3,465

Step-by-step explanation:

Given, Number of Democrats= 11

Number of Republicans = 7

Total persons in the group  =18

Number of congressman will be selected from the group = 8

Now , according to the combinations , the number of ways  of obtaining a committee with exactly 3 Democrats.= (3 Democrats , 5 Republicans)

$=^{11}C_3\times^7C_5\\\\=\dfrac{11!}{3!8!}\times\dfrac{7!}{5!2!}\\\\=\dfrac{11\times10\times9\times8!}{6\times8!}\times\dfrac{7\times6\times5!}{5!\times2}\\\\=165\times21=3465$

Hence, the number of ways of obtaining a committee with exactly 3 Democrats. = 3,465