@Anuj07 2??
@Faruq said:@vijay_chandola@ani4588We would like to form sets of 3 words from the letters contained in the set S={A, B, C, D, E, F, G, H, I} such that each word has 3 distinct letters and no two words share the same letter. How many different such sets of words can we have? a.60479 b.60478 c.60483 d.60480 e.60477Yeh toh question hi ni samajh aa raha
D answer hai kya ?
dividing them into 3 groups
9!/(3!*3!*3!*3!) gives 280
now arranging them
280*3!
now arranging individual letters
280*3!*3!
again
280*3!*3!*3!
gives 60480
dividing them into 3 groups
9!/(3!*3!*3!*3!) gives 280
now arranging them
280*3!
now arranging individual letters
280*3!*3!
again
280*3!*3!*3!
gives 60480
In how many ways can 12 different books can be divided equally among four persons ?
OE he dena coz i know the oa only 
@Anuj07 said:In how many ways can 12 different books can be divided equally among four persons ?OE he dena coz i know the oa only
@Faruq said:@vijay_chandola@ani4588We would like to form sets of 3 words from the letters contained in the set S={A, B, C, D, E, F, G, H, I} such that each word has 3 distinct letters and no two words share the same letter. How many different such sets of words can we have? a.60479 b.60478 c.60483 d.60480 e.60477Yeh toh question hi ni samajh aa raha
Is it 60480??
=(9P3*6P3*3P3)/3!
@Anuj07 said:In how many ways can 12 different books can be divided equally among four persons ?OE he dena coz i know the oa only
12!/(3!*3!*3!*3!*4!) equals to 15400
@Faruq
the first set of 3 things can be picked in
12C3 ways so remaining is 9 that can be picked in 9C3 ways and the next 6 can be picked up in 6C3 ways and the last 3 can be chosen in 3C3 ways
so total number of ways are
12C3 * 9C3 * 6C3 * 3C3 = 12! / ( 3!) ^ 4
12C3 ways so remaining is 9 that can be picked in 9C3 ways and the next 6 can be picked up in 6C3 ways and the last 3 can be chosen in 3C3 ways
so total number of ways are
12C3 * 9C3 * 6C3 * 3C3 = 12! / ( 3!) ^ 4
@Anuj07 said:In how many ways can 12 different books can be divided equally among four persons ?OE he dena coz i know the oa only
12!/(3!)^4
!!@Faruq said:@vijay_chandola @ani4588
We would like to form sets of 3 words from the letters contained in the set S={A, B, C, D, E, F, G, H, I} such that each word has 3 distinct letters and no two words share the same letter. How many different such sets of words can we have? a.60479 b.60478 c.60483 d.60480 e.60477
Yeh toh question hi ni samajh aa raha
9!/(3!)^4*(3!)^3 (DIVIDE AND THEN ARRANGE)
5 is in numerator, d looks most favoulable
@joyjitpal said:where did i go wrongplz explain
the first set of 3 things can be picked in
12C3 ways so remaining is 9 that can be picked in 9C3 ways and the next 6 can be picked up in 6C3 ways and the last 3 can be chosen in 3C3 ways
so total number of ways are
12C3 * 9C3 * 6C3 * 3C3 = 12! / ( 3!) ^ 4
12C3 ways so remaining is 9 that can be picked in 9C3 ways and the next 6 can be picked up in 6C3 ways and the last 3 can be chosen in 3C3 ways
so total number of ways are
12C3 * 9C3 * 6C3 * 3C3 = 12! / ( 3!) ^ 4
An ant is standing on the upper left corner of a 8 8 chess board, of which the four squares from the centre are removed. If the ant can only walk along the edges of the squares, and can only move down, or to the right, in how many different ways can it reach the lower right corner of the checkerboard?
@Anuj07 said:@ani4588Perfect solution !!How many words can be formed from the word letter INDIA
60? tuka hai... p&c; kuchh nahi aata....
@Anuj07 said:the first set of 3 things can be picked in12C3 ways so remaining is 9 that can be picked in 9C3 ways and the next 6 can be picked up in 6C3 ways and the last 3 can be chosen in 3C3 waysso total number of ways are12C3 * 9C3 * 6C3 * 3C3 = 12! / ( 3!) ^ 4
in this case you got to divide by 4!
as there are 4 groups having 3 each :)
as there are 4 groups having 3 each :)