CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[Aarav]

Hello All,

We are aware that the existing Newsletter software has problems that needs to be fixed. The technical guy in PG team is indisposed nowadays and Allwin is travelling and will be back in India after 8th. I promise to take this issue on top priority with PG team and get this rectified ASAP.

[billybolimeeaaaw]

The problem is similar to let's say 100 girls dancing on a floor one after another - in how many ways can they dance such that sita dances before gita who in turn dances before rita...
100!/3!
Same way, in this problem it 'll be 7!/3! = 840 option (4)

[howardroarkrock]

I subscribed to QQAD but havn't been getting the questions in my mailbox.
What is tha approx time it arrives in mailboxes?

cheers
Sayan

[tarkikpatel]

[quote=warrior;1144576]Quantitative Question # 034
------------------------------------------------------

In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?

(1) 736 (2) 768 (3) 792 (4) 840 (5) 876



hey aarav, i havent received the newsletter yet....i thot may be coz of the practice test...fortunately i logged in n found this questn lately....
well,, ppl hav already answered....sitter i guess
7 letters can be arranged in 7! ways
but since vowels are alike( of the same kind) we can consider them same...
so, 7!/3! * 1 coz of ascending order (only 1 casse we need ) = 840

[siddooba]

This one is easy ..

Number of ways of arranging JUPITER = 7!
However we need E I U in that order.
Number of ways of arranging the letters E,I,U = 3!

Therefore number of ways of arranging JUPITER with vowels in alphabetical order
= 7!/3!
= 7*6*5*4
= 840

Option (4)

Alternate Method:

Number of ways of Arranging JPTR = 4!
Now i have to insert E,I,U in between them.
Let
X1 = Number of letters before E
X2 = Number of letters between E and I
X3 = Number of letters between I and U
X4 = Number of letters after U

But X1+X2+X3+X4 = 4 (Since there are only 4 letters JPTR)

The number of solutions to an equation
a1+a2+....ar = k is (k+r-1)C(r-1)

Hence the number of ways of solving this is (4+4-1)C(4-1) = 7C3

Total ways of Arranging is = 7C3 * 4! = 7!/3! = 840

[sudeepdeb]

QQAD 034:

3 vowels , 4 consonants.
Since the sequence of vowels wont change, consider them as same.
Total Ways = 7! / 3! = 840.

Good that I saw a similar question few days back in official quant thread.

[m2shines]

Quantitative Question # 034
------------------------------------------------------

In how many ways can the letters of the word JUPITER be arranged in a row so that the vowels appear in alphabetic order?

(1) 736 (2) 768 (3) 792 (4) 840 (5) 876




we have 7 letters so if we stick E to first letter and I to 2nd letter
then no of ways = 5!
now of I is stick to 3rd position we have no of ways = 4!(4)
if we now move I to 4 to 5 to 6 to7 then we will get 3,2,1 ways of arranging U
so total ways with E at 1 = 5! + 4!(4+3+2+1)= 4!(10) +5!

now in same way if we stick E to 2nd postion and move I to 3 to last
total ways = 4!(4+3+2+1)= 4!(10)

Going in same way

and same way total ways = 5! + 4!(10) + 4!(10) + 4!(7) + 4!(3) + 4!
total = 5! + 4!(30) = 120 + 720 = 840

ANS (4).

My approach seems to be time taking but in P&C i feel 2 unsure about the answer that i always do some crosschecking.

[sanyo]

Consider the three vowels as same as only one combination is for them.

hence total number of arrangements = 7!/3!
= 7*6*5*4
= 840

[sanyo]

Consider the three vowels as same as only one combination is possible for them.

hence total number of arrangements = 7!/3!
= 7*6*5*4
= 840

[pnvishal]

Word JUPITER has three vowels E I U

In alphabatical order arrangement will be __ E __ I__U__

four gaps can contain four consonant. following cases will arise

case 1 One gap contains 4 letrs and others zero ,(4,0,0,0)

out of four gaps one can be choocsen in 4C1 = 4 ways

we can arrange the consonant in four ways 4!

so 4*4! = 96 ways

case 2 (1,3,0,0) arrangement

two gaps to arrange 1 and 3 letrs can be choosen in 4C2*2 ways

three letrs out of four consonant can be choosen and arrange in 4C3 *3! ways

so 12*4*3! = 288

case 3(2,2,0,0) arrangement

4C2*4C2*2!*2! = 144 ways for choosing gaps letrs and their arrangment

case 4(1,1,1,0)

4C1*3C1*4C2*2!*2! = 288 ways

case 5 (1,1,1,1)

4! = 24 ways


so total number of ways 96+288+144+288+24 = 840

So option 4.