penny has 25 dimes. she likes to arrange them into three piles, putting an odd number of dimes into each pile. in how many ways could she do this?

penny has 25 dimes. she likes to arrange them into three piles,

i think it is 78 too

Answer = 75

Explanation:-
As information provided in the question,

Total number of dimes that Penny has = 25 dimes

Total number of piles in which she likes to arrange the 25 dimes = 3 piles

Now,

It is given that she wants to put an odd number of dimes into each pile.

Thus,

Total number of possible pairs of three odd numbers that has a sum of 25 are as:

{1, 3, 21}, {1, 5, 19}, {1, 7, 17}, {1, 9, 15}, {1, 11, 13}

{3, 3, 19}, {3, 5, 17}, {3, 7, 15}, {3, 9, 13}, {3, 11, 11}

{5, 5, 15}, {5, 7, 13}, {5, 9, 11}

{7, 7, 11}, {7, 9, 9}

Now,

There are total 15 pairs in which 10 pairs have all three difinite elements and 5 pairs have two identical elements in each pair.

Therefore,

The number of ways in which we can arrange each pair which has all three distinct elements = 6 ways

And, The number of ways in which we can arrange each pair which has two identical elements elements = 3 ways

Therefore,

Total number of ways in which Penny can arrange all 25 dimes into three piles = 10 x 6 + 5 x 3

= 60 + 15

= 75 ways

7 ways

To find the number of ways Penny can arrange her dimes into three piles, each containing an odd number of dimes, we can use the concept of partitions.

First, let's consider how many odd numbers can be formed by adding three positive integers. We need to look at different cases for odd numbers:

Case 1: 1 + 1 + 1 = 3
Case 2: 3 + 1 + 1 = 5
Case 3: 3 + 3 + 1 = 7
Case 4: 5 + 1 + 1 = 7
Case 5: 5 + 3 + 1 = 9
Case 6: 5 + 5 + 1 = 11
Case 7: 7 + 1 + 1 = 9
Case 8: 7 + 3 + 1 = 11
Case 9: 7 + 5 + 1 = 13
Case 10: 7 + 5 + 3 = 15
Case 11: 9 + 1 + 1 = 11
Case 12: 9 + 3 + 1 = 13
Case 13: 9 + 5 + 1 = 15
Case 14: 9 + 5 + 3 = 17
Case 15: 9 + 7 + 1 = 17
Case 16: 9 + 7 + 3 = 19
Case 17: 9 + 7 + 5 = 21
Case 18: 11 + 1 + 1 = 13
Case 19: 11 + 3 + 1 = 15
Case 20: 11 + 5 + 1 = 17
Case 21: 11 + 5 + 3 = 19
Case 22: 11 + 7 + 1 = 19
Case 23: 11 + 7 + 3 = 21
Case 24: 11 + 7 + 5 = 23
Case 25: 11 + 9 + 1 = 21
Case 26: 11 + 9 + 3 = 23
Case 27: 11 + 9 + 5 = 25
Case 28: 11 + 9 + 7 = 27

Now, let's find the number of ways Penny can distribute her 25 dimes among the three piles.

We'll use the concept of stars and bars, where the stars represent the dimes and the bars represent the boundaries between the piles.

We have 25 dimes, so we need to distribute (place) 2 bars among the 25 dimes to create three piles. Using this concept, the number of ways to arrange the 25 dimes would be:

(N + K - 1) choose (K - 1)

In this case, N = 25 (number of dimes) and K = 3 (number of piles).

Therefore, the number of ways Penny can arrange her dimes into three piles, with each pile containing an odd number of dimes, is:

(25 + 3 - 1) choose (3 - 1) = 27 choose 2 = 351 possible ways.

So, Penny can arrange her dimes in 351 ways.