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# A person buys eight packets of TIDE detergent. Each packet contains one coupon, which bears one of the let-terms of the word TIDE. If he shows all the letters of the word TIDE, he gets one free packet. If he gets exactly one free packet, then the number of different possible combinations of the coupons isA.${}^{7}{{c}_{3}}$ B.${}^{8}{{c}_{4}}$ C.${}^{8}{{c}_{3}}$ D.${{4}^{4}}$

Last updated date: 25th Jun 2024
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Hint: Consider it as an 8 blank spaces $----$ as the 8 outcomes of the coupon, on each blank space we can assign 4 numbers T, I, D, E. Now If he gets exactly one free packet.
It means we want complete T I D E word to be formed only one time
So, we assign it like T I D E $--$ now we just have to perform permutation on blanks

Now like we have 8 coupons means 8 outcomes and, on each coupon, we can get 4 characters T, I, D, E but we get exactly one free packet, so it means complete TIDE word will be formed only one time
T I D E $--$
Now we can perform permutation on blank, each of the 4 blanks have 4 options T, I, D, E
so, we write it as ${{x}_{1}}+{{x}_{2}}+{{x}_{3}}+{{x}_{4}}=4......(1)$
here ${{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}$ denotes the number of T, I, D, E we take in blank spaces in different permutations,
where $1\le {{x}_{1}},{{x}_{2}},{{x}_{3}},{{x}_{4}}\le 8$
Then the required number of combinations of coupons is equivalent to the number of positive integral solutions of the equation (1),
Using formula
${{x}_{1}}+{{x}_{2}}+{{x}_{3}}.....+{{x}_{r}}=n$ total permutations will be ${}^{n+r-1}{{c}_{r-1}}$
On applying this formula to equation (1)
Our total permutation will be ${}^{4+4-1}{{c}_{4-1}}$
$\to {}^{7}{{c}_{3}}$
So, our final answer is ${}^{7}{{c}_{3}}$
So, the correct answer is “Option A”.

Note: We assigned like this T I D E $--$
But most of you might have doubt that why we didn’t count permutation of
T I D E word as $4!$ , because it doesn’t matter whether I got T I D E or E, D, I, T or I, T, D, E. In total we just want to make a complete word.