To solve this problem, we can use the combination formula:

nCr = n! / r!(n-r)!

where n is the total number of items, r is the number of items we want to select, and ! denotes factorial.

In this case, we want to select 4 letters from the word "EXAMINATION", which has 11 letters. So we have:

n = 11
r = 4

Substituting into the formula, we get:

11C4 = 11! / 4!(11-4)! = 11! / 4!7! = (11x10x9x8) / (4x3x2x1) = 3300 / 24 = 136.

Therefore, the correct answer is option A, 136.

Ham pehle 4 blanks banayengey...
_ _ _ _

Case 1 lenge;

Kyunki repetition hai kucch letters ki toh ham un letters ko packet mei band kar denge and us packet ko 1 simple entity consider karenge. hamaare paas 3 packets banenge i.e. [AA], [II], [NN].

(_ _ ) (_ _)
Now, Since letters ke packets bane hai toh obviously gaps ke bhi packets banenge. Toh gaps ke ban gaye two packets. Because 2 spaces par (_ _) same hi letter aayega.

Ab arrangement toh karna nahi. Toh obviously COMBO lagaayenge.

3 packets ki choices hai
2 packets ki places bharni hai.

³C2

³C2 = 3*2/2*1 = 3ways

Case 2,
_ _ _ _

2 gaps mei same letter, baaki 2 mei different letter.

yaani (_ _) ( _) ( _)

We have now, 3 choices.
These 3 choices can be filled by 11 letters.
But 3 packets are there that repeat. And inn 3 packets ko ham siraf aur siraf 1 hi jagah par place kar sakte hai. So ³C1 = 3ways

Now,
2 places bachi hai. letters reh gaye 8 (including pairs) but since ham 1 packet of letter already place kar chuke hai. So remaining letters are 8-1 = 7.

⁷C2 = 7*6/2*1 = 21ways

Abh question ye hai ki is case nei add kare ya multiply?
toh dekho, since ye case tabhi poora hoga jab ye 4 letter word banke taiyaar ho jaayega. Since aisa nahi hai ki pehle ³C1 se hamaara letter ban gaya ho. ya siraf ⁷C2 se hamaare 4 letters ki space bhar gayi ho. because they can't exist independently or complete the 4 blank spaces independently toh we MULTIPLY.
i.e = 21*3 = 63ways
Case3,

(_) ( _) ( _) (_)

4 jagah hai.
11 letters hai. lekin 3 repeat hai toh 3 packets mei band hai. Yaani letters bache 8.

8 choices hai. 4 jagah hai.
⁸C4 = 8*7*6*5/4*3*2*1 = 70ways.

TOTAL ANSWER = 70+ 63+ 3 = 136ways.

Hope you got this. if you still have any doubt. so comment it.
Thank you.