# Difficult Permutation-Combination Solved QuestionAptitude Discussion

 Q. In how many ways can a selection of 5 letters be made out of  $5A$'s, $4B$'s, $3C$'s, $2D$'s and $1E$'s?
 ✖ A. $70$ ✔ B. $71$ ✖ C. ${^{15}C_5}$ ✖ D. ${^{15}P_5}$

Solution:
Option(B) is correct

Number of ways of selecting 5 different letters:

$= {^5C_5} = 1$ way

Number of ways to select 2 similar and 3 different letter:

$={^4C_1} ×{^4C_3}=16$

Number of ways of selecting 2 similar + 2 more similar letter and 1 different letter:

$= {^4C_2}×{^3C_1}= 18$

Number of ways to select 3 similar and 2 different letter:

$= {^3C_1}×{^4C_2}= 18$

Number of ways to select 3 similar and another 2 other similar:

$= {^3C_1}×{^3C_1}=9$

Number of ways to select 4 similar and 1 different letter:

$= {^2C_1}×{^4C_1}=8$

ways of selecting 5 similar letters:

$= 1$

total ways: $= 1+16+18+18+9+8+1= \textbf{71}$

Alternately

A more detailed solution is as follows:

Table below can be scrolled horizontally

5 A’s
A A A A A
4 B’s
B B B B
3 C’s
C C C
2 D’s
D D
1 E
E
Sub
Total

SC1
5 same
0 diff.

${^1C_1}$
$=1$

NA

NA

NA

NA

1

SC2
4 same
1 diff.

$1×{^4C_1}$
$=4$

$1×{^4C_1}$
$=4$

NA

NA

NA

8

SC3
3 same
2 diff. each

$1×{^4C_2}$
$=6$

$1×{^4C_2}$
$=6$

$1×{^4C_2}$
$=6$

NA

NA

18

SC4
3 same
2 other same

$1×{^3C_1}$
$=3$

$1×{^3C_1}$
$=3$

$1×{^3C_1}$
$=3$

NA

NA

9

SC5
2 same
3 diff. each

$1×{^4C_3}$
$=4$

$1×{^4C_3}$
$=4$

$1×{^4C_3}$
$=4$

$1×{^4C_3}$
$=4$

NA

16

SC6
2 same
2 other same
1 diff.

$1×{^3C_1}×{^3C_1}$
$=9$

$1×{^2C_1}×{^3C_1}$
$=6$

$1×{^1C_1}×{^3C_1}$
$=3$

NA

NA

18

SC7
all diff.

${^5C_5}$
$=1$

1

Sum =

71

The solution can be structured in 7 'selection conditions' (SC 1 to SC 7)

SC1:

All five letters are the same

Only possible for $A = 1$ letter

⇒ Number of ways:

$= {^1C_1} = 1$ way -------- (1)

SC2:

Only 4 Letters are the same (Only possible with $A$ and $B$) and one is different (1 out of 4 remaining, after $A$ or $B$ is selected)

⇒ Number of ways:

$= 1×{^4C_1} + 1×{^4C_1} = 8$ ways -------- (2)

(Same as ${^2C_1}×{^4C_1}= 8$)

SC3:

Only 3 Letters are the same (Possible with $A, B$ and $C$) and two are diff.( 2 out of 4 remaining, after $A$ or $B$ or $C$ is selected)

⇒ Number of ways:

$= 1×{^4C_2}+1×{^4C_2}+1×{^4C_2} = 18$ ways -------- (3)

(same as ${^3C_1}×{^4C_2} = 18$)

SC4:

3 Letters are the same (Possible with $A, B$ and $C$) and two others are same (Possible with $A, B, C$ and $D ⇒ 2$ letters of 1 letter type from 3 remaining letter types)

⇒ Number of ways:

$= 1×{^3C_1}+1×{^3C_1}+1×{^3C_1} = 9$ ways -------- (4)

(Same as ${^3C_1}×{^3C_1} = 9$)

SC5:

2 Letters are the same (Possible with $A, B, C$ and $D$) and three are diff (same logic as above)

⇒ Number of ways:

$= 1×{^4C_3}+1×{^4C_3}+1×{^4C_3}+1×{^4C_3} = 16$ ways -------- (5)

(same as ${^4C_1}×{^4C_3} = 16$)

SC6:

2 Letters are the same (Possible with $A, B, C$ and $D$) and 2 other Letters are the same (Possible with $A, B, C$ and $D$)and one different letter (same logic as above)

⇒ Number of ways:

$= 1×{^3C_1}×{^3C_1} +1×{^2C_1}×{^3C_1} +1×{^1C_1}×{^3C_1} = 18$ ways -------- (6)

(Same as ${^4C_2}×{^3C_1}= 18$)

[SC6 is a little tricky because we need to avoid repetition as shown above]

SC7:

All 5 Letters are diff (Possible only with $A, B, C, D$ and $E$)

⇒ Number of ways $= {^5C_5} = 1$ way -------- (7)

Adding the number of ways from the 7 SCs:

$= (1) + (2) + (3) + (4) + (5) + (6) + (7)$

$= 1+8+18+9+16+18+1$

$= \textbf{71}$

## (1) Comment(s)

Vinoth Kumar N
()

Sir, for the explanation of this question. Searched the whole internet but everyone reproduced the same answer without any logic. Superb very happy for the detailed answer. Now I'm confident that I can answer questions of this type. Thank you, sir.