Mashrur's billiard ball choosing

We want an even number as our sum. And we have $3$ ways of doing that.

Case $1$ : ( even+even+even+even=even )

There are $7$ even numbers and we want to choose $4$ . So we can do that in $7C4=35$ ways.

Case $2$ : (odd+odd+odd+odd=even)

There are $8$ odd numbers and we want to choose $4$ . So we can do that in $8C4=70$ ways.

Case $3$ : (odd+odd+even+even=even) If we want to pick $2$ odd numbers and $2$ even numbers, we can do that in $8C2 \times 7C2=588$ ways.

So the total number of ways for Mashrur to take the balls is $35+70+588=\boxed{693}$

There are $7$ balls with even numbers and $8$ balls with odd numbers. In order to the sum of all $4$ balls be even, there are three ways...

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• Case 1 : All the balls are numbered even. Since there are $7$ even numbered balls and we need to choose $4$ , we have $\dbinom{7}{4}$ ways to do so.

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• Case 2 : All the balls are numbered odd. Since there are $8$ odd numbered balls and we need to choose $4$ , we have $\dbinom{8}{4}$ ways to do so.

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• Case 3 : Two of the balls are numbered even, other two are numbered odd. As mentioned before, there are $7$ even numbered balls and to choose $2$ out of them, we have $\dbinom{7}{2}$ ways to do so. Then, there are $8$ odd numbered balls and to choose $2$ out of them, we have $\dbinom{8}{2}$ ways to do so. Hence we have $\dbinom{7}{2} \cdot \dbinom{8}{2}$ ways to do so.

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In total, we get:

$\dbinom{7}{4}+\dbinom{8}{4}+\dbinom{7}{2} \cdot \dbinom{8}{2} = \fbox{693}~\text{total ways}$ .

This is easy. First, we list out all possible cases which will cause the 4 numbers on the balls have a even sum. They are

1. 2 even-numbered and 2-odd numbered balls

2. all of them are odd-numbered

3. all of them are even-numbered.

Now, we calculate the number of possibilities case by case.

In Case 1, we have $^7C_2 \times ^8C_2 = 21\times28 = 588$ ways.

In Case 2, we have $^8C_4=70$ ways.

In Case 3, we have $^7C_4=35$ ways.

So, in total, we have $\boxed{693}$ possibilities.

Note that instead of using permutation, we need to use combination, since $1+2+3+4$ is equal to $4+3+2+1$ .

x+y+z+k=even,implies either all four even,all four odd,2 even and 2 odd,since all these ways this even sum can be achieved....................so there are 7 even and 8 odd numbers so.......so no of ways 7c4+8c4+8c2*7c2...............

there are 7 even numbered and 8 odd numbered balls. now the cases to pick the balls are either all 4 even or 2 even,2 odd or 4 odd. so the ways are respectively $\frac{7!}{3!*4!} , \frac{(7!)*(8!)}{(2!*5!)*(2!*6!)} , \frac{8!}{4!*4!}$ . So the sum of the three is 693.

Within the given balls, there are $7$ even-numbered balls and $8$ odd-numbered balls.

Given the laws of parity, the sum of four different natural numbers can be even in three different cases:

1. All four numbers are even

2. Two numbers are odd, two numbers are even

3. All four numbers are odd

Summing up these cases, we get 7C4 + 7C2 x 8C2 + 8C4 = 693

(8C4)+(8C2)(7C2)+(7C4)=693

There are 8 odds and 7 even number. The possible sum of 4 number so that its become the even if there are 4 odds, 2odds + 2 even, and 4 even. So we get 4C8 + 2C6 x 2C7 + 4C7 = 35 + 588 + 70 = 693.

First, let's generalize all the possible scenarios that a 4-number combination will have an even number sum. Note that the order of the numbers is not important.

Let $e$ be the set of all even integers and $o$ be the set of all odd integers. Using their definition:

$e = \left\{ {2k \mid k \in \mathbb{Z}} \right\}$ $o = \left\{ {(2k + 1) \mid k \in \mathbb{Z}} \right\}$

Going back to the given, there are $8$ odd numbers from 1 to 15: 1, 3, 5, 7, 9, 11, 13, & 15.

Also, there are $7$ even numbers from 1 to 15: 2, 4, 6, 8, 10, 12, & 14.

We are looking for an even sum from the 4-number combinations. There are 3 possible scenarios to get that sum:

1. EVEN + EVEN + EVEN + EVEN
2. ODD + ODD + ODD + ODD
3. EVEN + EVEN + ODD + ODD

For the first scenario. there are $\binom{7}{4} = 35$ possible ways of getting an all-even, 4-numbered combination.

Secondly, there are $\binom{8}{4} = 70$ ways of getting an all-odd combination.

Lastly, for the last one, since there are $\binom{7}{2} = 21$ ways of getting 2 even-number combination from the set, and $\binom{8}{2} = 28$ ways of getting 2 odd-number combination from the same set, then there are $21\cdot{28} = 588$ ways of getting 2 even-numbered and 2 odd-numbered for a 4-number combination from the given set.

To sum it up, there are:

$35 + 70 + 588 = \boxed{693}$

ways of getting 4-number combination with an even number sum.

from 1 to 15, there are 8 odd numbers and 7 even numbers

8C4 = 70

7C4 = 35

8C2 x 7C2 = 588

70 + 35 + 588 = 693

first of all we must take the possibilities of all combinations resulting in an even sum.. (odd+odd+odd+odd),(even+even+even+even) and (odd+odd+even+even) There are 7 even and 8 odd nos ...... so we must add the combination of taking 4 from 8 odd, taking 4 from 7 even and the product of the combination of taking 2 from 8 odd and 2 from 7 even
so, 8C4+7C4+8C2*7C2=693

There are 3 cases such that 4 numbers on the balls have a even sum. They are

1.even+odd+even+odd=even

2.odd+odd+odd+odd=even

3.even+even+even+even=even

So

In Case 1, we have $^7C_2 \times ^8C_2 = 21\times28 = 588$ ways.

In Case 2, we have $^8C_4=70$ ways.

In Case 3, we have $^7C_4=35$ ways.

So, we have $\boxed{693}$ ways.

Balls could be all even, all odd, two even and two odd: C(8,4)+C(7,2)*C(8,2)+C(7,4) = 693

Here we have 7 even numbers and 8 odd numbers.there are three ways to make an even sum. no.1# 7C4=35 no.2#7C2*8C2=588 no.3#8C4=70 in total=35+588+70=693.

693 think like this....1. 4 even 2. 2 odd,2even 3. 4odd........ now use combination formula for desired result....thanks

Consider $3$ cases:

$A)$ $4$ even numbers:

There are $7$ even numbers from $1$ to $15$ . The combinations are $\frac {7!}{4!\cdot3!} = 35$ .

$B)$ $4$ odd numbers:

There are $8$ odd numbers from $1$ to $15$ . The number of combinations are $\frac {8!}{4!\cdot4!} = 70$ .

$C)$ $2$ even numbers and $2$ odd numbers:

Well, there are $7$ even numbers from $1$ to $15$ . The different combinations, choosing $2$ of them, are $\frac {7!}{2!\cdot5!} = 21$ . Similarly, with the odd numbers, we'll get $28$ . That means that for each of the $21$ pairs of even numbers, there are $28$ pairs of odd numbers. In other words, $21\cdot28 = 588$ .

Finally, the answer is $70 + 35 + 588 = \boxed {693}$ .