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If N = p! + q! + r! is a two digit prime number. Then how many such value of N exist

(Q). If N = p! + q! + r! is a two digit prime number. Then how many such value of N exist.
Solution:-

0! = 1 3! = 61! = 1 4! = 242! = 2 5! = 120 ✖Now make combination from these values

∴ 0! + 3! + 4! = 31            &   1! + 3! + 4! = 31
                          ↓                                            ↓
                       prime                                   prime
&  no other cases
Hence 2 such values of N exists. Answer

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