Okay, The math problem involves factorials
457!
____
454 !
2007-09-09 10:38:31 · 4 answers · asked by J.E Lane 2 in Science & Mathematics ➔ Mathematics
457! / 454! = 457 * 456 * 455 * (454! / 454!)
= 457 * 456 * 455 * (1)
= 94,818,360
2007-09-09 10:50:08 · answer #1 · answered by euclidjr 2 · 0⤊ 0⤋
457! / 454!
= 457 * 456 * 455 * 454! / 454! (note: 454! / 454! cancel out)
= 457 * 456 * 455
2007-09-12 23:19:59 · answer #2 · answered by Pakyuol 7 · 0⤊ 0⤋
457! / 454!
(457 * 456 * 455 * 454!) / 454!
The 454! cancels out
So you have
(457 * 456 * 455) = 94,818,360
.
2007-09-09 10:50:31 · answer #3 · answered by Robert L 7 · 0⤊ 0⤋
The triangle could be anybody that includes a, and a minimum of a million element from the backside row, so which you have a million achievable first element (A), 4 achievable 2d factors (Any of the backside row) and seven achievable third factors. So if F is your 2d element, B,C,D,E,G,H, and that i could be your third, if G is your 2d, B,C,D,E,H, and that i could be your third, if H is your 2d, B,C,D,E, and that i could be your third, and if I is your 2d, B,C,D, and E could be your third. so which you have 7 with F, 6 with G, 5 with H, and four with I. so in case you upload those 4 numbers, 4+5+6+7=22 achievable triangles. desire that helps.
2016-10-18 11:27:34 · answer #4 · answered by ? 4 · 0⤊ 0⤋