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 
${^1C_1}$ 
NA 
NA 
NA 
NA 
1 
SC2 
$1×{^4C_1}$ 
$1×{^4C_1}$ 
NA 
NA 
NA 
8 
SC3 
$1×{^4C_2}$ 
$1×{^4C_2}$ 
$1×{^4C_2}$ 
NA 
NA 
18 
SC4 
$1×{^3C_1}$ 
$1×{^3C_1}$ 
$1×{^3C_1}$ 
NA 
NA 
9 
SC5 
$1×{^4C_3}$ 
$1×{^4C_3}$ 
$1×{^4C_3}$ 
$1×{^4C_3}$ 
NA 
16 
SC6 
$1×{^3C_1}×{^3C_1}$ 
$1×{^2C_1}×{^3C_1}$ 
$1×{^1C_1}×{^3C_1}$ 
NA 
NA 
18 
SC7 




${^5C_5}$ 
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:
$= (1) + (2) + (3) + (4) + (5) + (6) + (7)$
$= 1+8+18+9+16+18+1$
$= \textbf{71}$
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.