Holi problem

The Python program below finds that Harsh doesn't get wet $1048576 = 4^{10}$ different ways, and at least one of them didn't get wet $4662625 = 5^3 \times 37301$ different ways. Therefore, $a = 3$ , $c = 37301$ , $d = 4$ , $e = 10$ , and $(a + c) - (d + e) = \boxed{37290}$ .

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import string
digs = string.digits + string.ascii_letters

def int2base(x, base):
    if x < 0:
        sign = -1
    elif x == 0:
        return digs[0]
    else:
        sign = 1
    x *= sign
    digits = []
    while x:
        digits.append(digs[int(x % base)])
        x = int(x / base)
    if sign < 0:
        digits.append('-')
    digits.reverse()
    return ''.join(digits)

Htotal = 0
Atotal = 0
for a in range(0, 5 ** 10):
    seq = int2base(a, 5).zfill(10)
    H = False
    A = False
    if "0" not in seq:
        H = True
        A = True
    if "1" not in seq:
        A = True
    if "2" not in seq:
        A = True
    if "3" not in seq:
        A = True
    if "4" not in seq:
        A = True
    if H:
        Htotal += 1
    if A:
        Atotal += 1
print(Htotal, Atotal)

By Inclusion-Exclusion, the number of ways where everyone gets wet is $\sum_{k = 0}^5 (-1)^k \binom{5}{k} (5 - k)^{10} = 5103000,$ so the number of ways where at least one person remains dry is $5^{10} - 5103000 = 4662625 = 5^3 \cdot 37301.$ Thus, the answer is $3 + 37301 - (4 + 10) = 37290$ .

LET 10 FRIENDS THAT THROW WATER BE REPRESENTED BY F1 AND OTHERS BE REPRESENTED BY F2.

Harsh didn't want to get wet.Which means that only 4 of the friends(F2) can get wet by 10 other friends(F1) so number of ways will be $4^{10}$ . Therefore d=4 ; e = 10.

Total number of ways in which friends (F1) can make others wet = $5^{10}$ .

Selection of any 5 friends(F2) can be done in 10!/5!5! and this 5 friends(F1) can throw water in 5! ways when none of them throw at a friend(F2) twice.The rest of the five friend(F1) can throw water at any of them and therefore total number of ways is $5^{5}$ .

Hence total number of ways in which everyone(F2) gets wet is[ $5^{5}$ ] [10!/5!5!]. Total number of ways in which at least one doesn't get wet is = Total number of ways - total n.o of ways in which everyone gets wet. => $5^{10}$ - $5^{5}$ 10!/5!5!.

= $5^{5}$ ( $5^{5}$ -10!/5!5!) = $5^{5}$ (3125-252) = $5^{5}$ (2873) Therefore a=5 and c=2873. = (a+c)-(d+e) =(2873+5)-(10+4) =2878-14=2864