An algebra class of 21 students must send 5 students to meet with the principal. How many different groups of students could be formed from this class?
Ahh I totally forgot how to do this. Can someone explain?
2007-12-30 13:20:43 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
How many different groups of 5 students*
2007-12-30 13:21:34 · update #1
21C5= 20349 different groups.
You choose [nCr] instead of [nPr] because the order in which the students are chosen DOESN'T matter.
2007-12-30 13:24:19 · answer #1 · answered by ¿ /\/ 馬 ? 7 · 4⤊ 1⤋
21!/(21-5)!= 21*20*19*18*17=2441880
you have 21 choices for the first student and 1 less to chose from for each student after that.
but since the group does not care what order they are in,
you must devide by 5! to get rid of the groups with the same kids in different order 5!=120
so the final answer = 20349
2007-12-30 13:40:33 · answer #2 · answered by saejin 4 · 0⤊ 0⤋
Just Multiply 21 x 5= 105~
Best Wishes!
2007-12-30 13:42:02 · answer #3 · answered by ♥~Kirsten~♥ 3 · 0⤊ 1⤋
Basic permutation calculation. You have 21 to choose from for the first choice. That leaves 20 for the second, 19 for the third, 18 for fourth, and 17 choices for the fifth spot.
Multiply all these choices to get the total number of permutations (combinations) of students in this group of 5:
21*20*19*18*17 = 2,441,880
Who knew you could have so many choices with such a small group! :-)
2007-12-30 13:35:01 · answer #4 · answered by Charles M 6 · 0⤊ 2⤋
21!/((21-5)!*5!)=20349
combinations problem given by formula :
n=total amout
r=amount chosen
n!/ ( (n-r)! * r!)
other way of explaining:
the first child can by anyone of 21 kids
2nd can be any of 20
3rd can be any of 19
4th can be any of 18
5th can be any of 17
thus 21*20*19*18*17 which is also 21!/(21-5)!
but with 5 kids, they can be ordered different but still be the same group thus u divide the answer from the last equation by 5!
so:
21!/(21-5)! then divide again by 5!
same as the original formula 21!/ ( (21-5)!*5!) =20349
2007-12-30 13:26:03 · answer #5 · answered by X W 2 · 3⤊ 1⤋
Just multiply 21 and 5, which equals 105 different groups of students can be formed by this class.... Best of luck!!!
2007-12-30 13:23:46 · answer #6 · answered by amber211 2 · 2⤊ 5⤋
isn't it 21 to the power of 5?
2007-12-30 13:24:30 · answer #7 · answered by Anonymous · 1⤊ 6⤋
by multiply, 21x5=105
by combination=21c5=
=21x20x19x18x17
------------------------
5x4x3x2x1
=20,349
by permutation
21p5= 21!
-------------
5!(21-5)
= 21x20x91x18x17x16
---------------------------------
5!x 16!
=488,376
2007-12-30 13:59:26 · answer #8 · answered by Keethu 2 · 0⤊ 1⤋
i think you just multiply 21 and 5
...not sure though
2007-12-30 13:25:27 · answer #9 · answered by Justy 2 · 0⤊ 6⤋