3consider the set x = { 1 , 2 , 3 , 4 , 5 } and suppose you have two holes. also suppose that you have 10 pigeons: the 2 -element subsets of x . can you put these 10 pigeons into the two holes in a way that there is no 3 -element subset s = { a , b , c } ⊂ x for which all pigeons from s go in the same hole? then answer the same question if x = { 1 , 2 , 3 , 4 , 5 , 6 } with 15 = c ( 6 , 2 ) pigeons.

answer: 140

step-by-step explanation:

so just add all then it will be your answer but i am not pretty sur eif it is but hope this i am here to people

answer: 48

just remember to use pemdas:

parentheses

expressions

multiplication

division

addition

subtraction

3,6 and 7.

step-by-step explanation:

a) unit place of 37 is 7

so the unit place of cube of 37 will be

7³= 343=3

the answer is 3.

b) unit place of 56 is 6, so the unit place of cube of 56 will be

6³= 216=6

the answer is 6.

c) unit place of 133 is 3, so the unit place of cube of 133 will be

3³= 27= 7.

the answer is 7.

add the numbers under frequency to find the total number:

5 + 8 +7 + 4 + 0 +1 = 25

the answer is 25.