Assume that the committee consists of 7 Republicans
and 6 Democrats. A subcommittee consisting of 5 people is to
be selected.
(1) How many such subcommittees are possible if each subcommittee must
contain exactly 3 Republicans and 2 Democrats?
(2) How many such subcommittees are possible if each subcommittee must
contain at least 1 and no more than 2 Democrats?
and 6 Democrats. A subcommittee consisting of 5 people is to
be selected.
(1) How many such subcommittees are possible if each subcommittee must
contain exactly 3 Republicans and 2 Democrats?
(2) How many such subcommittees are possible if each subcommittee must
contain at least 1 and no more than 2 Democrats?
-
Formula for permutations: nPk = n!/(n-k)!
Formula for combinations: nCk = n!/(n-k)!k!
1.) Say our 7 republicans' names are r1, r2, r3, r4, r5, r6, and r7
Our 6 democrats' names are d1, d2, d3, d4, d5, and d6
So, a working configuration would be
r1 r2 r3 d1 d2
r1 r2 r4 d1 d3
r1 r2 r5 d1 d4
etc... See the pattern? 7C3 is the combinations of r's and 6C2 is the combinations for d's. How do we combine them?
7C3 is 35 and 6C2 is 15. If you think of the example above in two parts, the r's and the d's, then there are 7C3 ways to arrange the R's for every single way to arrange the D's. So we multiply them to get there are
525 ways to arrange 7 Republicans and 6 Democrats into a subcommittee with exactly 3 Republicans and 2 Democrats.
2.) At least 1 and no more than 2, that sounds like find out number where there are 4 R and 1 D and add it to 3R and 2D.
So by the same logic,
7C4 * 6C1 + 7C3 * 6C2 = answer to #2
35 * 6 + 35 * 15 = answer to #2
210 + 525 =
735 possibilities.
Hopefully I helped! And hopefully I did it right :P
Formula for combinations: nCk = n!/(n-k)!k!
1.) Say our 7 republicans' names are r1, r2, r3, r4, r5, r6, and r7
Our 6 democrats' names are d1, d2, d3, d4, d5, and d6
So, a working configuration would be
r1 r2 r3 d1 d2
r1 r2 r4 d1 d3
r1 r2 r5 d1 d4
etc... See the pattern? 7C3 is the combinations of r's and 6C2 is the combinations for d's. How do we combine them?
7C3 is 35 and 6C2 is 15. If you think of the example above in two parts, the r's and the d's, then there are 7C3 ways to arrange the R's for every single way to arrange the D's. So we multiply them to get there are
525 ways to arrange 7 Republicans and 6 Democrats into a subcommittee with exactly 3 Republicans and 2 Democrats.
2.) At least 1 and no more than 2, that sounds like find out number where there are 4 R and 1 D and add it to 3R and 2D.
So by the same logic,
7C4 * 6C1 + 7C3 * 6C2 = answer to #2
35 * 6 + 35 * 15 = answer to #2
210 + 525 =
735 possibilities.
Hopefully I helped! And hopefully I did it right :P