4C3+5C3+6C3+7C3+8C3+9C3 will be here nCr = n!/[r!*(n-r)!] and ! is factorial for example 6! = 6

M4maths Previous Puzzles

28 Nov 2012 combination

4C3+5C3+6C3+7C3+8C3+9C3 will be
here nCr = n!/[r!*(n-r)!]
and ! is factorial
for example 6! = 6*5*4*3*2*1 and 5! = 5*4*3**2*1

OPtion

1) 134
2) 209
3) 311
4) 349
5) 396
6) 456
7) 549
8) 679
9) 999
10) none of these

Solution

it is important to notice that nCr+nC(r-1)=(n+1)Cr
and 4C4 = 1
so 4C3+5C3+6C3+7C3+8C3+9C3
= 4C4+4C3+5C3+6C3+7C3+8C3+9C3-4C4
= 10C4-1
= 210-1 = 209

Correct Option: 2 Best Solution (12)

birenparekh   11 years AGO

simple as,
4c3=4,5c3=10,6c3=20,7c3=35,8c3=56,9c3=84
so add all=209

Like? Yes (7) | No (1)  

FOLSHIYA SATHYARAJ   11 years AGO

4c3=4
5c3=10
6c3=20
7c3=35
8c3=56
9c3=84
so
4
+10
+20
+35
+56
+84=209

Like? Yes (2) | No   

Ashokkumar   11 years AGO

Ans:
= 4c1+5c2+6c3+7c4+8c5+9c6
= 4+10+20+35+56+84
= 209

Like? Yes (2) | No   

Jitendra Kumar Kaushik   11 years AGO

4+10+20+35+56+84=209

Like? Yes (2) | No   

Awadhesh Kumar   11 years AGO

nCr= nCn-r

Like? Yes (1) | No   

Sourav   11 years AGO

calculating each factorial we get this sum

Like? Yes (1) | No   

syed ashwaq ahmed   11 years AGO

apply the given formula to the above and solve it !!

Like? Yes (1) | No   

Balakumaran N.A   11 years AGO

4C3 = 4C1 =4;
5C3 = 5*4*3 / 3*2*1 = 10; like this for all...

then, 4+10+20+35+56+84 = 209.

Like? Yes (1) | No   

Chandolu.Naga Vijay Seshank   11 years AGO

by using the given formula
4!/3!+5!/(3!*2!)+6!/(3!*3!)+7!/(3!*4!)+8!/(3!*5!)+9!/(3!*6!)=4+10+20+35+56+84=209

Like? Yes (1) | No   

Amit Kumar   11 years AGO

(4c3=4)+(5c3=5*4)+(6c3=6*5*4)+...so on=209

Like? Yes (1) | No   

Ratan Lal Jat   11 years AGO

4+10+20+35+56+84=209

Like? Yes (1) | No   

rapolu srikanth   11 years AGO

4+10+20+35+56+84=209
So ans:2

Like? Yes (1) | No