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The ways of selecting 4 letters from the word EXAMINATION is
- a)136
- b)130
- c)125
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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The ways of selecting 4 letters from the word EXAMINATION isa)136b)130...
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The ways of selecting 4 letters from the word EXAMINATION isa)136b)130...
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.
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.
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The ways of selecting 4 letters from the word EXAMINATION isa)136b)130...
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.
_ _ _ _
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.
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The ways of selecting 4 letters from the word EXAMINATION isa)136b)130c)125d)none of theseCorrect answer is option 'A'. Can you explain this answer? for CA Foundation 2025 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The ways of selecting 4 letters from the word EXAMINATION isa)136b)130c)125d)none of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for CA Foundation 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ways of selecting 4 letters from the word EXAMINATION isa)136b)130c)125d)none of theseCorrect answer is option 'A'. Can you explain this answer?.
The ways of selecting 4 letters from the word EXAMINATION isa)136b)130c)125d)none of theseCorrect answer is option 'A'. Can you explain this answer? for CA Foundation 2025 is part of CA Foundation preparation. The Question and answers have been prepared according to the CA Foundation exam syllabus. Information about The ways of selecting 4 letters from the word EXAMINATION isa)136b)130c)125d)none of theseCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for CA Foundation 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The ways of selecting 4 letters from the word EXAMINATION isa)136b)130c)125d)none of theseCorrect answer is option 'A'. Can you explain this answer?.
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