MATHS EXPERTS PLS HELP!!!NOW!!

[CIE xams]

pls help me in solving thesse permutations and cmbinations questions

1.In how many ways can 8 similar balls be placed in three diffrent boxes when each box contains at least 1 ball...
ans: 21
2.if 8 balls to b selected from 3 balls of different colours, balls of each colour being availbale unlimited no. of times.total no. of selection is:
45
3.no. of ways in which 4 letters of the word MATHEMATICS can be arranged:
ans:2454
4. no. of squares on a chessboard:
ans: 204(i know this cud be done as 1^2+2^2+3^2+....8^2, but pls explain how to do it in terms of permutations or combinations)
PLS HELP FAAAAST ANYONE PLSSSSSSS....

 

[Jinosupreme]

For q1, I listed out all the ways.
Make it into box A,B,C.
Then set one of it to be constant of 1, so if I see Box A to be one all the time, ,it will be
A. B. C.
1. 1. 6.
1. 2. 5.
1. 3. 4.
1. 4. 5.
1. 5. 2.
1. 1. 6.

First outcome is 6.


Then i set box B to be constant with all 1 ball
A. B. C.
1. 1. 6. (INVALID AS IT IS REPITITION OF THE PREVIOUS OUTCOME, HENCE CANCEL THIS OUT)
2. 1. 5.
3. 1. 4.
4. 1. 3.
5. 1. 2.
6. 1. 1.

This outcome is 5 (cause i minus the REPITITION)

Then do the same to BOX C, set it constant as 1, then u will get total of 4 outcomes.

Now, put second condition which is one box must be 2 as constant
So if I put Box A for constant of 2 balls
A. B. C.
2. 1. 5. (CANCEL THIS OUT AS IT IS A REPITITON OF ABOVE)
2. 2. 4.
2. 3. 3.
2. 4. 2.
2. 5. 1. (CANCEL THIS CAUSE IT IS ALSO A REPITITON)

Outcome is 3.
Then do the same for box B and C
Then add all the outcomes
6+5+4+3+2+1=21

(I know this is not the proper way to do it, but I hope I helped)

 

[Jinosupreme]

Question 3.

Must use selection to select first then only you can arrange it.
MATHEMATICS
Object=11
Number of M = 2
Number of A = 2
Number of T = 2
Number of Different alphabets in it = 8 (M A T H E I C S)

So first set your condition.
Condition 1: out of 4 letters, 2 letters are the same
Hence,
Selection1:
3C1 X 7C2 = 63

Explanation:
3C1 because among the pairs of M, A, and T, you choose 1.
7C2 because if you chose M for 3C1, then u left 7 different alphabets to choose from!

Arrangement 1:
63 X (4!/2!) = 756

Explanation:
After u select, you get 63.
Then u arrange it by multiply 4!/2! Because there's 2 alphabets repeated in the arrangement

Now you done with 1st condition move on to 2nd condition
Condition 2: out of 4 letters, there are 2 pairs of alphabets repeating.

Selection:
3C2 = 3

Explanation:
Because among M A T, u choose 2 pairs to make 4 letters

Arrangement:
3 x (4!/(2! x 2!)) = 18

Explanation:
U got 3 selections then u arrange it, multiply (4!/(2! x 2!)) cause there's 2 pairs of same alphabet.

Condtition3: No repeating among 4 letters

Selection:
8C4 = 70

Explanation:
There are 8 different alphabets in the word, so u choose 4.

Arrangement :
70 X 4! = 1680

Explanation:
No repetition, hence multiply 4!

Then add all the arrangement:

756 + 18 + 1680 = 2454