A music store wants to display 3 identical keyboards, 2 identical trumpets, and 2 identical guitars in its store window. How many distinguishable displays are possible?
2007-01-20 14:47:58 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
i'm first (i also like jokes)
So there are places n= 3+2+2 =7; first consider they are all different instruments, then number of variants P=n! Now consider keyboards are identical, then P=n!/3!;
The same for the rest; thus P=7!/3!/2!/2! =210;
Or! 3 identical keyboards in 7 places have number of variants Ck= 7*6*5/3!, number of vacant places left being = 7-3=4; two identical trumpets in 4 places have number of variants Ct= 4*3/2!, number of vacant places left being = 4-2=2, quite enough for 2 identical guitars, Cg=1; thus P=Ck*Ct*Cg=210; typical bostonsoxnm.
2007-01-20 14:52:01 · answer #1 · answered by Anonymous · 2⤊ 3⤋
The 2 guitars can occupy the 7 positions in 7*6/2 or 14 ways
The same applies to the trumpets.
The 3 keyboards have 7*6*5/2/3 = 35 ways
14*14*35 = 6,860 different displays
2007-01-20 15:17:23 · answer #2 · answered by Helmut 7 · 0⤊ 1⤋
There are 7!/(3!2!2!) = 210 ways.
To see why, notice that if the instruments are all different there are 7! displays, but for a given display you can permute each type of instrument and still have the same display, you need to divide by the number of ways of permuting each type of instrument. 3! ways for the keyboards, 2! ways for the trumpets and 2! ways for the guitars.
2007-01-20 15:26:03 · answer #3 · answered by Phineas Bogg 6 · 1⤊ 0⤋
18
2007-01-20 14:54:49 · answer #4 · answered by melinda_trvn 2 · 0⤊ 1⤋
This stuff is kinda like factoring but a bit more into it.
Just take 3x2x2 and there's the answer.
2007-01-20 14:58:06 · answer #5 · answered by Anonymous · 0⤊ 1⤋
3x2x2=18
2007-01-20 15:13:50 · answer #6 · answered by CheeseLord 3 · 0⤊ 1⤋
Infinite . . .
You did say first ANSWER!
2007-01-20 14:52:58 · answer #7 · answered by Say What? 5 · 1⤊ 3⤋