= 5 × 4 × 3 × 2 × 1 = 120 words so 25th word will definitely start with A

And after A next letter is C so words starting with AC are = 4!(number of letters left in JACKED other than A, C) = 4 × 3 × 2 × 1 = 24. So 25th word can’t start with AC.

And after AC in dictionary combination, to start words will be with AD and so 25th word will start with AD which is ADCEJK (letters other than AD are arranged in alphabetical order because dictionary follows this pattern)

= 6 × 5 × 4 × 3 × 2 × 1 = 720 words so 55th word will definitely start with A.

After A next letter is D so words starting with AD are = 5!(number of letters left in PRANKED other than AD) = 5 × 4 × 3 × 2 × 1 = 120 so 55th word will definitely start with AD.

After AD next letter is E so words starting with ADE are = 4!(number of letters left in PRANKED other than ADE) = 4 × 3 × 2 × 1 = 24 so 55th word is not included in it

And after combination of ADE next combination is ADK so words forming with ADK = 4!(number of letters left in PRANKED other than ADK) = 4 × 3 × 2 × 1 = 24 so 24 + 24 = 48 so 55th word is not included in it.

After combination of ADK next combination is ADN so words forming with ADN = 4!(number of letters left in PRANKED other than ADN) = 4 × 3 × 2 × 1 = 24 so 24 + 24 +24 = 72 so 55th word is included in it.

So ADNE will be the next combination and with it 3! = 3 × 2 × 1 = 6 words can form so 48 + 6 = 54

So after ADNE next combination is ADNK and with it 3! = 3 × 2 × 1 = 6 words can form so ADNKEPR is the 55th word.

n(A ∩ B ∩ C) = 2, n(A ∩ B) = 3, n(A ∩ C) = 3 and n(B ∩ C) = 3

⇒ Number of people in committee A and B but not in C = 3 – 2 = 1

⇒ Number of people in committee A and C but not in B = 3 – 2 = 1

So minimum number of people in committee A = 1 + 1 + 2 = 4

Number of ways of selecting r objects from n objects = [latex]^{n}{c}_{r}[/latex]

Number of ways to select 2 out of [latex] 8 = ^{8}{c}_{2} = \frac {8!}{(2! × 6!)} = 28[/latex]

Number of ways of selecting r objects from n objects = [latex]^{n}{c}_{r}[/latex]

Number of ways to select 6 out of 9 choices = [latex] ^{9}{c}_{6} = \frac {9!}{(6! × 3!)} = 84[/latex] ways

© 2013 - 2023 SPLessons.com