Find the number of ways in which an arrangement of four letters can be can be made from the letters (contd)?

Question

of the word -PROPORTION.

if ur doing the question by taking cases like 1)3alike 1diff 2)2alike 2alike 3)2alike 2diff 4)all diff. then plz explain clearly why we take 'PP' or 'OOO' as a single element, without dividing the answer by 2! and 3! respectively.
And also that, why in case 2) we don not consider 'OO' out of 'OOO' as a possibility of being 2 alike.
thanks a lot for ur help....

if u are not able to understand the two questions i've put up in the in details section, then plz do the question in UR WAY, but then, plz explain every step clearly.....

Answer 1

A very interesting problem.

I personally agree with Northstar's idea to classify the problem into four different cases. However, my answers to cases 2, 3 and 4 are different from his.

Case 1. All four different letters: c(6,4)4! = 360
Reason: Pick 4 out of 6 distinct letters and do permutation.

Case 2. Two letters the same, the other two unique: c(3,1)c(5,2)4!/2! = 360
Reason: Pick one pair from 3 available two-letter pairs and 2 from the remaining 5 distinct letters, and do permutation.

Case 3. Two pairs of two letters the same: c(3, 2)4!/(2!2!) = 18
Reason: Pick 2 pairs from 3 availabletwo-letter pairs and do permutation.

Case 4. Three letters the same, one unique:c(3,3)c(5,1)4!/3! = 20
Reason: Pick all three O letters c(3,3) and 1 from the remaining 5 distinct letters, and do permutation.

Total = 360+360+18+20 = 758

--------------------------
Puzzling,
Your following statement is not right.
"In terms of arranging you have C(4,2) = 6 ways to place the two different letters, and the two alike letters just fall into place."

An easy way to place 4 letters with two identical ones is to do permutation and then divide the permutation by 2: 4!/2! = 12

Or you can think about this way. Focus on the two distinct letters. You have four places for the first one and three remaining places for the second one, then the two identical letters fall into the other two places. So this is 4x3 = 12.

Answer 2

Start with 3 alike and 1 different. The only choice for the 3 alike is the letter O. Then you have 4 places to put the remaining letter of which there are 5 letters (P, R, T, I or N).

So the answer for that part is: 1 x 4 x 5 = 20 ways

C(4,1) x C(5,1) = 20

OOOP OOOR OOOT OOOI OOON
OOPO OORO OOTO OOIO OONO
OPOO OROO OTOO OIOO ONOO
POOO ROOO TOOO IOOO NOOO

Next try 2 alike 2 alike. The choices for 2 alike are O, P or R. You have C(3, 2) choices for the pairing. Then you have to figure the ways to arrange the two pairs of letters. You only have to consider one pair of letters. There are two letters to place in 4 spots, so you have C(4,2) positions to chose.

So that is 3 x 6 = 18

C(3,2) x C(4,2) = 18

OOPP OORR PPRR
OPOP OROR PRPR
OPPO ORRO PRRP
POOP ROOR RPPR
POPO RORO RPRP
PPOO RROO RRPP

The next case is 2 alike and two different.

Here you have C(3, 1) = 3 choices for the double letter (P, R, O)
Then you have C(5, 2) = 10 choices for the different letters.

In terms of arranging you have C(4,2) = 6 ways to place the two different letters, and the two alike letters just fall into place.

So 3 x 10 x 6 = 180 ways

I won't list them however.

Finally you have all different. There are 6 choices for the first, 5 for the second, 4 for the third and 3 for the last. Alternatively think of picking the four different letters from the six possible choices, C(6,4) = 15 and then arranging them 4! = 24 ways. That is C(6,4) x 4! = 360 ways

Adding it all up:
20 + 18 + 360 + 180
= 578 ways

Answer 3

PROPORTION = OOOPPRRINT = 3O,2P,2R,I,N,T

Ok to start with here's the basic result for number of permutations of four letters with repetition, for the separate cases:

[All different] 4! = 24

[2 alike, 2 different] 4!/2! = 4*3 = 12

[3 alike, 1different] 4!/3! = 4

[2 alike, 2 alike] 4!/(2!2!) = 4*3/2 = 6

>plz explain clearly why we take 'PP' or 'OOO' as a single element, without dividing the answer by 2! and 3! respectively.

It's wrong to think of 'PP' or 'OOO as a "single element", think of them instead as equivalent permutations which are self-similar in 2! and 3! ways respectively.
i.e. P1P2P3 would be equivalent to P3P1P2 etc.

> And also that, why in case 2) 2alike 2alike we don not consider 'OO' out of 'OOO' as a possibility of being 2 alike.
Because that case is covered separately under 3 alike.

So finally, no. of permutations in which arrangements of four letters from 'PROPORTION'
(10 letters, 6 unique letters, 3O,2P,2R,I,N,T)

This is going to be, for each case:
(the number of permutations in which you can choose the letters involved) * (the number of ways in which those 4 letters can be arranged, i.e. one of the results for the cases above)

So:
[ 3O + 1 other from 5 ] 5_C_1 * 4_C_1 = 5 * 4 = 20
[ 2O/2P/2R+ 2 other unique letters from the remaining 5]
(3_C_1 * 5_C_2) * 4_C_2 = 3 * 5*4/2 * 4*3/2 = 180
[ 2 alike, 2 alike] 3_C_2 * 4!/(2!2!) = 3 * 6 = 18
[ 4 different ] 6_C_4 * 4! = 6!/(4!2!) * 4! = 6!/2 = 360
Total 20+180+360+18=578

Answer 4

The word "proportion" can be dissected as shown below.

i
n
ooo
pp
rr
t

The possibilities are for choosing four letters are:

All four different letters
Two letters the same, the other two unique
Two pairs of two letters the same
Three letters the same, one unique

Each of the combinations have a certain number of arrangements.

So the number of different arrangements are:

(4!)(6C4) + (12)(3)(4C2) + (4)(3C2) + (4)(1)(5C3)
= 24*15 + 12*3*6 + 4*3 + 4*1*10
= 360 + 216 + 12 + 40 = 628

Answer 5

here p=2,r=2,o=3
total number of arrangements of four letters=10p4/2!2!3!
10!/4!2!2!3!(ans)

Answer 6

Your question does not make sense.